Leetcode
秤和砣
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Best Time to Buy and Sell Stock
class Solution { public: int maxProfit(vector &prices) { // Note: The Solution object is instantiated only once and is reused by each test case. int solution = 0;转载 2014-04-28 15:10:46 · 421 阅读 · 0 评论 -
Plus One
Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. class Solution { publ转载 2014-04-28 15:02:05 · 346 阅读 · 0 评论 -
Minimum Path Sum
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at转载 2014-04-29 15:14:01 · 295 阅读 · 0 评论 -
Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. // RECURSIVE VERSION:转载 2014-04-28 15:47:39 · 348 阅读 · 0 评论 -
Permutations
// NON-SWAP VERSION: class Solution { public: vector > solution; vector singleSolution; void helper(vector & num, vector & visited){ if (singleSolution.size()==num.size转载 2014-04-28 13:54:18 · 334 阅读 · 0 评论 -
Generate Parentheses
void helper(vector & solution, string & singleSolution, int numofleft, int numofright, int n){ if (singleSolution.size()==(2*n)) solution.push_back(singleSolutio转载 2014-04-28 14:43:52 · 321 阅读 · 0 评论 -
Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7},转载 2014-04-28 14:19:42 · 203 阅读 · 0 评论 -
Gray Code
class Solution { public: vector grayCode(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function vector result; result.pu转载 2014-04-28 13:10:12 · 300 阅读 · 0 评论
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