摘要：
Code:O(n):class Solution {public: vector twoSum(vector &numbers, int target) { vector res; map mapping; int n=numbers.size(); for (int i=0; i twoSum(vector &numbers, int target) { vector res; int n=numbers.size(); for(int i=0;i<n;i++) fo... 阅读全文
摘要：
Code:class Solution {public: int maxDep; vector buf; // pack numbers into an array vector> res; // pack arrays into an result void dfs(int dep, vector &valid, vector &num){ if(dep==maxDep){ res.push_back(buf); return; } for(int i=0; ... 阅读全文
摘要：
Code:class Solution {public: int singleNumber(int A[], int n) { // Note: The Solution object is instantiated only once and is reused by each test case. int res = 0; int bit,i,j; for(i=0;i>i)&1==1) bit = (bit+1)%3; } if(bit!=... 阅读全文
摘要：
Analysis:1. Sort the vector;2. Set i, j, k and traverse;3. Start from i, j is after i, and k is the last element;4. If theTrick: By increasing j and decreasing k, to reduce the complexity from n^3 to n^2.class Solution {public: int threeSumClosest(vector &num, int target) { sort(num.begin(... 阅读全文
摘要：
Definition for singly-linked list./** * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */Method 1: (Double-Pointer)class Solution {public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ListNode *p1 = NULL; ListNode *... 阅读全文
摘要：
Objective:Find out the lowest point to buy-in and find out the a following hightest point to sell-out. And the maximal profit is equivalent to the max difference between the highest point and lowest point.Before solving this problem, suggest referring to the "maximum subarray" first. Then 阅读全文
摘要：
Code:class Solution {public: int maxDep; vectorbuf; vector> res; void dfs(int dep, vector &valid, vector &num){ if(dep==maxDep){ res.push_back(buf); return; } for(int i=0;i> permuteUnique(vector &num) { buf.clear(); res.clear(); ... 阅读全文
摘要：
题目：输入一个整型数组，数据元素有正数也有负数，求元素组合成连续子数组之和最大的子数组，要求时间复杂度为O(n)。例如：输入的数组为1, -2,3, 10, -4, 7, 2, -5，最大和的连续子数组为3, 10, -4, 7, 2，其最大和为18。背景：本题最初为2005年浙江大学计算机系考研题的最后一道程序设计题，在2006年里包括google在内的很多知名公司都把本题当作面试题。由于本题在网络中广为流传，本题也顺利成为2006年程序员面试题中经典中的经典。分析：如果不考虑时间复杂度，我们可以枚举出所有子数组并求出他们的和。不过非常遗憾的是，由于长度为n的数组有O(n2)个子数组（即：n 阅读全文