This document is to summarize the solutions of LeetCode problems.

• 3Sum: sort the array first, then make three pointers - i to head, j to tail and k to pivot. if (a[i] + a[k] + a[j]) < 0, i++ else j--.
• 3SumCloset: similar to 3Sum.
• 4Sum: compute the sum of each pair of two elements in the array frist. for each twoSum i, check whether another twoSum j exist, where i + j = target.
• AddBinary: simple problem.
• AddTwoNumbers: simple problem.
• Anagrams: simple #hashtable problem.
• BalancedBinaryTree: simple #balance #tree problem.
• BestTimetoBuyandSellStock: simple problem.
• BestTimetoBuyandSellStockII: simple problem.
• BestTimetoBuyandSellStockIII: tricky #dp problem. divide and conquer. find the max profit of 0..i and i+1...n, each of them is in just one transaction. get the i for optimal profit.
• BinaryTreeInorderTraversal: simple #tree binarytree problem. recursive method is trival. iterative method is as follows. one stack recs is used to store each node. if the left node is NULL, pop up the top node i in the stack recs, and then traversal the node i->right.
• BinaryTreeLevelOrderTraversal: simple #tree #binarytree #bst problem. #bst the tree level by level.
• BinaryTreeLevelOrderTraversalII: simple #tree #binarytree #bst problem. the same method used in BinaryTreeLevelOrderTraversal, and then reverse the result.
• BinaryTreeMaximumPathSum: simple #tree #binarytree problem. for each node i, max(leftPathSum+i->val, rightPathSum+i->val, leftPathSum+rightPathSum+i->val) and return max(leftPathSum+i->val, rightPathSum+i->val) to parent.
• BinaryiTreeZigzagLevelOrderTraversal: simple #tree #binarytree problem. use odd and even level to control whether reverse the vector.
• Candy: tricky problem. scan from beginning, if ratings[i] > ratings[i-1] num[i] = num[i-1] + 1. It can assign the candy to each increasing interval. Then, scan from the end, if num[i] is already assigned by a number, if ratings[i] > ratings[i+1], then i is at the peak, make sure num[i] > num[i+1], otherwise, num[i] = num[i+1] + 1. if num[i] is not assigned yet, make it to be 1.
• ClimbingStairs: simple #dp problem. steps[i] = steps[i-1] +steps[i-2].
• CloneGraph: #hashtable problem. use a #hash to store the nodes already created. Every time, first check whether the node is already created.
• CombinationSum: simple knapsack problem. #dp. The numbers can be reused. Scan from the beginning. Use unordered_map<int, <vector<int> > path to store the previous number that can reach this number. Use recussive method to retrieve the path.
• CombinationSumII: simple knapsack problem. #dp. The numbers could not be reused. Scan from the end. Use unordered_map<int, <vector<int> > path to store the previous number's index that can reach this number. Use recussive method to retrieve the path.
• Combinations: simple #dfs problem.
• ConstructBinaryTreefromInorderandPostorderTraversal: simple #tree #binarytree problem. The last elemeent in postorder traversal is the root node which can be used to split the inorder traversal.
• ConstructBinaryTreefromPreorderandInorderTraversal: simple #tree #binarytree problem. The first elemeent in preorder traversal is the root node which can be used to split the inorder traversal.
• ContainerWithMostWater: tricky problem. Two pointers, i to the frist, j to the end. Compute max(res, min(height[i], height[j])*(j-i)), Then i++ if height[i] < height[j], otherwise, j--.
• ConvertSortedArraytoBinarySearchTree: simple #tree #binarytree #binarysearch #recussive problem. Use element in the middle as the root each time. and build left subtree and right subtree using left section and right section in the array.
• ConvertSortedListtoBinarySearchTree: simple #tree #binarytree #linkedlist #recussive problem. The same method used in ConvertSortedArraytoBinarySearchTree, but need a function to compute the length of list and find the element in list by index.
• CopyListwithRandomPointer: #hashtable. The same method used in CloneGraph. A #hashtable is used to store the node already created.
• CountandSay: simple problem. Just simple use the rule described in the problem to generate the string.
• DecodeWays: simple #dp problem. opt[i] = opt[i-2] + opt[i-1] if s[i-1]..s[i] can be decoded, otherwise, opt[i] = opt[i-1].
• DistinctSubsequences: simple #dp problem. opt[i][j] represents the number of distinct subsequences for T[0]..T[i-1] and S[0]..S[j-1]. opt[i+1][j+1] = opt[i][j] + opt[i+1][j] if T[i] == S[j], otherwise, opt[i+1][j+1] = opt[i+1][j].
• DivideTwoIntegers: trick #bitshift problem. b << 1 until b is larger than a with i times shift and record the result after each shift in inc[i]. Then, a -= inc[i] and res += 1 << i until a <= 0 or i < 0.
• EditDistance: classic #dp problem.
• FirstMissingPositive: trick problem. swap elements to put positive number i into the position i-1. scan from the beginning, find the first position i where a[i] != i+1, then return i+1.
• FlattenBinaryTreetoLinkedList: #tree #binarytree problem. for each node i, find the right most node j for i->left, then make j->right = i->right, i->right = i->left and i->left = NULL.
• GasStation: tricky problem. scan all gas stations and compute total += gas[i] - cost[i] and tank += gas[i] - cost[i]. if tank < 0, mark i and set tank = 0. after done, if total > 0, then return (i+1) % gas.size() else -1. The reason why (i+1) % gas.size() could be a starting point is that for every previous point k, tank(k) = tank(k-1) + gas[k] - cost[k], where tank(k-1) >= 0, it could not make tank(i) = tank(i-1) + gas[i] - cost[i] >= 0, then it would also not make tank(i) >= 0 if tank(k) = gas[k] - cost[k], which makes k to be the starting point.
• GenerateParentheses: #dfs #recusive problem. generate ( frist, then genrate ).
• GrayCode: convert a binary to the gray code by GrayCode(i) = (i>>1)^i.
• ImplementstrStr(): simple problem.
• InsertInterval: scan the whole intervals, each time get the smallest for newInterval.start and the largest for newInterval.end.
• InsertionSortList: #linkedlist problem. split the list, n1->n2->n3->n4->null, into two part, a sorted part, n0->n1->n2->null, and an unsorted part, n3->n4->null. Each time go through the sorted part to find a position for the element would be inserted.
• IntegertoRoman: simple problem. check from the largest number represents in Roman.
• InterleavingString: tricky #dp problem. opt[i+1][j+1] represents whether s1[0]...s1[i] and s2[0]...s2[j] can be interleaved to construct s3. check s1[i] == s3[i+j+1] and s2[j] == s3[i+j+1] to make opt[i+1][j+1] = opt[i][j+1] || opt[i+1][j].
• JumpGame: simple problem. each time check wehther this position can be reached by previuos position.
• JumpGameII: find the further position for each step.
• LargestRectangleinHistogram: tricky problem. push each area into the stack, until the current height is lower than that of the top element in the stack. Then, pop up the areas in the stack and compute them for each top element in the stack whose height is higher or equals to current height. Then, push the new area with current height and the new width into the stack again.
• LengthofLastWord: simple problem.
• LetterCombinationsofaPhoneNumber: simple problem.
• LinkedListCycle: classic problem. two nodes, one moves forward one step, another moves forward two step, check wether thes two nodes would be overlapped.
• LinkedListCycleII: classic problem. after detect the cycle using the method in LinkedListCycle. Go through from head and the intersected node, if they meet, then return that node.
• LongestCommonPrefix: hash the prefix of each string. #hashtable
• LongestConsecutiveSequence: tricky problem. put all elements into a hash first. iterate the elements in the hash, and check the length of left and right consecutive in the hash. #hashtable
• LongestPalindromicSubstring: scan the index of the string, check whether palindrome when the index k is in the middle.
• LongestSubstringWithoutRepeatingCharacters: tricky problem. use prev[] to keep the nearest index where s[prev[i]] = s[i]. if it is the first occurence of the element, prev[i] = -1. scan the elements, if prev[end] >= begin, begin = prev[end] + 1.
• LongestValidParentheses: tricky problem. more practice. use a struct to store the value s[i] and the index i. push '(' into stack. each time find a match for ')', compute the length by i - top(stack).index after the matched '(' has been popped.
• LRUCache: #lru #hashtable problem. a struct unordered_map<int, pair(list<int>::iterator, int)> is used for cache, list<int> for keep the order of data.
• MaximalRectangle: tricky problem. use height[] to indicate the number of consecutive 1 appearred in each column to current line. and then, the problem would be reduced to solve LargestRectangleinHistogram.
• MaximumDepthofBinaryTree: simple problem.
• MaximumSubarray: simple problem.
• MedianofTwoSortedArrays: tricky problem. use divide and conquer. if A[mid] < B[mid], then the value should not in A[0]...A[mid]. Similarly, B[mid] < A[mid], then the value should not in B[0]...A[mid].
• MergeIntervals: similar to InsertInterval.
• MergekSortedLists: simply #linkedlist problem.
• MergeSortedArray: do it in-space. pointer k to the last index of A's capacity. pointer i to the last index of A's actual values, pointer j to the last index of B. each time determine whether A[i] or B[j] will be assigned to A[k].
• MergeTwoSortedLists: simple linkedlist problem.
• MinimumDepthofBinaryTree: simple #binarytree #tree problem.
• MinimumPathSum: simple #dp problem.
• MinimumWindowSubstring: tricky problem. two pointers - head and tail. tail would be increased, until find a window (a #hashtable is used to make such a decision). Then, shrink the window by increasing head.
• MultiplyStrings: multiply manually. more practice.
• N-Queens: simple #dfs problem.
• N-QueensII: simple #dfs problem.
• NextPermutation: tricky problem. First, start from the end of the sequence, find the first index i where an increasing sequence exists. Then start from the end of the sequence again, find the first index j, where a[i] < a[j] and i < j. swap a[i] and a[j] and reverse(a.begin()+i+1, a.end()).
• PalindromeNumber: simple problem.
• PalindromePartitioning: partition the string into two parts, if the first part is palindrome, then continue. PalindromePartitioningII: tricky #dp problem. #dp to determine whether a substring, s[i]..s[j], palindrome - recs[i][j] = s[i] == s[j] && recs[i+1][j-1]. use #dp to determine whether a cut is required - opt[j] = min(opt[j], opt[i]+1), if s[i+1]..[j] is palindrome.
• PartitionList : simple #linkedlist problem. use one #linkedlist to keep tracking the first part, use another #linkedlist to keep tracking the second part. concatenate these two parts in the end.
• Pascal'sTriangle: simple problem.
• Pascal'sTriangleII: simple problem. a[i] += a[i+1], a.insert(a.begin(), 1).
• PathSum: simple #tree #binarytree problem.
• PathSumII: simple #tree #birnarytree problem.
• PermutationSequence: tricky problem. The number in the 1st position, should repeate (n-1)! times. So, we could use k/(n-1)! to calcualte which number should be located in 1st position. Then, k %= (n-1)!. Use the same method to cacluate the number in 2nd, 3rd, ..., positions.
• Permutations: simple problem. swap a[i] and a[j] where i <= j and recussively do it on i+1.
• PermutationsII: tricky problem. Each iteration, use the method in NextPermutation to generate the next permutation for num and then add it into the result. It wwill end when num comes to be the same value as that is at the beginning.
• PlusOne: simple problem.
• PopulatingNextRightPointersinEachNode: simple #tree #binarytree problem. node->left->next = node->right ? node->right : NULL. node->right->next = node->next->left.
• PopulatingNextRightPointersinEachNode: simple #tree #binarytree problem. node->left->next = node->right ? node->right : getNext(node). node->right->next = getNext(node). In getNext(node), get the next sibling by iterating all node = node->next and return node->left or node->right. After one node finished, go further to node->right first, since it can build the next pointer which may be used in the left node.
• RecoverBinarySearchTree: #tree binarytree problem. inorder traversal the tree. first found prev->val > node(j)->val then prev would be the first node in wrong position. the second time found prev->val > node(j)->val then j woudl be the second node.
• Pow(x, n): tricky #bitshift problem. say, 3^5 = (3^1)*(3^4) = (3^(2^0))*(3^(2^2)) and 5 in binary is 101. make n in binary, if the bit i in n is 1, then we can have x^(2^i).
• RegularExpressionMatching: tricky problem. one pointer i to the first string s1, one pointer j to the second string s2. Each time if s2[j+1] == '*', we need to check isMatch(s1[i]...s1[n], s2[j+2]...s2[m]) and increase i until s1[i] != s2[j]. Then, we need to check isMatch(s1[i]...s1[n], s2[j+2]...s2[m]) for next segment. if s2[j+1] != '*', if s1[i] == s2[j] then check isMatch(s1[i+1]...s1[n], s2[j+1]...s2[m]), otherwise, return false. '.' would be treated as the same to any s1[i].
• RemoveDuplicatesfromSortedArray: simple problem.
• RemoveDuplicatesfromSortedArrayII: simple problem.
• RemoveDuplicatesfromSortedList: simple #linkedlist problem. Try with double pointer.
• RemoveDuplicatesfromSortedListII: simple #linkedlist problem. Try with double pointer.
• RemoveElement: simple problem. Two pointers, i to head, j to tail. each time A[i] == elem, swap(A[i], A[j--]), otherwise i++.
• RemoveNthNodeFromEndofList: simple #linkedlist problem. Try with double pointer.
• ReverseInteger: simple problem.
• ReverseLinkedListII: simple problem. try 'double pointer'. Each time change the tail node to the head.
• ReverseNodesink-Group: simple problem. similar to ReverseLinkedListII. try double pointer.
• ReorderList: #linkedlist problem. use emplace_back() to push all elements into a vector. Use two pointers i, j to add elements from i or j.
• RestoreIPAddresses: simple #dfs problem. check 0 <= each segment <= 255.
• RomantoInteger: simple problem.
• RotateImage: tricky problem. need more practice. rotate by layer. think about the swap.
• RotateList: simple #linkedlist problem.
• SameTree: simple #tree #binarytree problem.
• ScrambleString: tricky problem. #dp. for each s1[i1]..s1[j1] and s2[i2]..s2[j2], check isScramble(s1[i1], s1[i1+i], s2[i2], s2[i2+i]) && isScramble(s1[i1+i+1], s1[j1], s2[i2+i+1], s2[j2]) and isScramble(s1[i1], s1[i1+i], s2[j2-i+1], s2[j2]) && isScramble(s1[i1+i+1], s1[j1], s2[i2], s2[j2-i]).
• Searcha2DMatrix: simple problem. #binarysearch.
• SearchforaRange: simple problem. #binarysearch.
• SearchinRotatedSortedArray: tricky #binarysearch problem. more practice. if A[mid] >= A[begin], then check A[mid] > target && target >= A[begin]; otherwise, check A[mid] > target || target >= A[begin].
• SearchinRotatedSortedArrayII: tricky #binarysearch problem. more practice. similar to SearchinRotatedSortedArray, except if A[mid] == A[begin], begin++.
• SearchInsertPosition: simple #binarysearch problem.
• SetMatrixZeroes: tricky problem. if matrix[i][j] == 0, set matrix[i][0] = 0 and matrix[0][j] = 0. Then, check matrix[i][0] and matrix[0][i] to set 0 on the whole row or column.
• SimplifyPath: consider the rule. more practice.
• SingleNumber: tricky problem. XOR to all the elements.
• SingleNumberII: tricky problem. #bitproblem. calculate the occurrences of each number in each bit and mod 3.
• SortColors: simple problem. three pointers, i, k, j to 0, 1, 2, respectively. move k to determine swap(A[i], A[k]) or swap(A[k], A[j]) or k++.
• SpiralMatrix: tricky problem. four pointer beginX, endX, beginY, endY. each time move beginY++, endX--, endY--, beginX++.
• SpiralMatrixII: the same method as that of SpiralMatrix.
• Sqrt(x): #binarysearch. use long long to avoid overflow, since multiplication may cause overflow.
• StringtoInteger(atoi): consider all the possible inputs.
• Subsets: #dfs problem. each time the element can be added to the subset or not.
• SubsetsII: #dfs problem. for each duplicated elements, just find the end position of the duplicated element, and add them once.
• SubstringwithConcatenationofAllWords: tricky problem. #hashtable to record all strings in L. for each position in S, check whether it is a good start using another #hashtable to record the strings matched in S.
• SudokuSolver: find each '.' and try 1-9 on it.
• SumRoottoLeafNumbers: simple problem. #tree #binarytree.
• SurroundedRegions: #dfs. first check all 'O' on the margin and mark them and 'O's surrounded them. then modify all 'O's which are not marked.
• SwapNodesinPairs: #linkedlist. try double pointers on it.
• SymmetricTree: #tree #binarytree problem. try recursive and iterative method. recursive method: each time check two nodes - i, j - and check isSymmetric(i->left, j->right) and isSymmetric(i->right, j->left). iterative method, check by level.
• TextJustification: consider the number of space.
• TrappingRainWater: How mach water can be contained in index i is determined by the heighest one on the left and right.
• Triangle: calculate buttom-up.
• TwoSum: #hashtable to record the position.
• UniqueBinarySearchTrees: tricky #tree #binarytreeproblem. iterate the root nodeithenresult += numTrees(left(i)) * numTrees(right(i))`.
• UniqueBinarySearchTreesII: The same method as that of UniqueBinarySearchTrees. recursively generate the tree by iterating the node as root. Generate all leftChildTrees and rightChildTrees, and then generate them by making the node as root.
• UniquePaths: simple #recursion problem.
• UniquePathsII: simple #recursion problem. when Obstacles exists, not go through with it.
• ValidateBinarySearchTree: #tree #binarytree problem. check lower bound and upper bound for each node.
• ValidNumber: need more practice.
• ValidPalindrome: two pointers - i, j - to head and tail. just check when s[i] and s[j] are alphanumeric.
• ValidParentheses: simple #stack problem.
• ValidSudoku: check whether valid for each row, column and square.
• WildcardMatching: trick #dp problem. #recursive will be TLE. opt[i][j] represents whether s[0]...s[i] will match p[0]...p[j].
• WordBreak: #dp problem. opt[j] represents whether s[0]...s[j] could be combined in dictionary. opt[j] = opt[i] if substring(i+1, j) exists in dictionary.
• WordBreakII: #dp problem. same method in WordBreak. Need a #hashtable to record the path.
• WordLadder: #bfs problem. iterate a to z for each position of each word.
• WordLadderII: #bfs problem. need more practice. need to handle overlapping when add path. use another #hashtable to store which values have been reached before this level.
• WordSearch: #dfs problem.
• ZigZagConversion: j += 2*nRows-2, k = j + 2 * (nRows - i - 1).

The following problems come from #leetcode blogs.

• StringReorderDistanceApart: #gready problem. each time check for the word with highest frequency first.
• Multiplicationofnumbers: tricky problem. use left to be the multiplication of all elements on the left of i. right to be the multiplication of all elements on the right of i.
• Rotatinganarrayinplace: tricky problem. rotate the string triple times.