1. Two Sum★

题目内容：Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

题目分析：Array 　　Hash table

题目要求输入的是和队列的数值相关的值，输出的适合队列的索引相关的值，那么可以将队列存储在Map当中，Map中的Key是队列中的数值，Value是队列中与数值相关的索引。注意Map的定义语句：

Map <Integer, Integer> map = new HashMap <Integer, Integer>();

题目代码：

时间： 2024-12-18 05:52:29

1. Two Sum★的相关文章

LeetCode OJ - Sum Root to Leaf Numbers

这道题也很简单,只要把二叉树按照宽度优先的策略遍历一遍,就可以解决问题,采用递归方法越是简单. 下面是AC代码: 1 /** 2 * Sum Root to Leaf Numbers 3 * 采用递归的方法,宽度遍历 4 */ 5 int result=0; 6 public int sumNumbers(TreeNode root){ 7 8 bFSearch(root,0); 9 return result; 10 } 11 private void bFSearch(TreeNode ro

129. Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum of all root-to-leaf numbers. For example, 1 / 2 3 T

Leetcode 494 Target Sum 动态规划背包+滚动数据

这是一道水题,作为没有货的水货楼主如是说. 题意:已知一个数组nums {a1,a2,a3,.....,an}(其中0<ai <=1000(1<=k<=n, n<=20))和一个数S c1a1c2a2c3a3......cnan = S, 其中ci(1<=i<=n)可以在加号和减号之中任选. 求有多少种{c1,c2,c3,...,cn}的排列能使上述等式成立. 例如: 输入:nums is [1, 1, 1, 1, 1], S is 3. 输出 : 5符合要求5种

31.SUM() 函数

SUM() 函数 SUM 函数返回数值列的总数(总额). SQL SUM() 语法 SELECT SUM(column_name) FROM table_name SQL SUM() 实例我们拥有下面这个 "Orders" 表: O_Id OrderDate OrderPrice Customer 1 2008/12/29 1000 Bush 2 2008/11/23 1600 Carter 3 2008/10/05 700 Bush 4 2008/09/28 300 Bush 5

1305 Pairwise Sum and Divide

基准时间限制:1 秒空间限制:131072 KB 分值: 5 难度:1级算法题有这样一段程序,fun会对整数数组A进行求值,其中Floor表示向下取整: fun(A) sum = 0 for i = 1 to A.length for j = i+1 to A.length sum = sum + Floor((A[i]+A[j])/(A[i]*A[j])) return sum 给出数组A,由你来计算fun(A)的结果.例如:A = {1, 4, 1},fun(A) = [5/4] + [

Java [Leetcode 303]Range Sum Query - Immutable

题目描述: Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You may assume that the ar

LeetCode 303. Range Sum Query - Immutable

求数组nums[i,j]的和思路:另开一sum数组,sum[i]为nums[0,i]的和,所以nums[i,j] = sum[j] - sum[i-1] 1 class NumArray { 2 public: 3 vector<int> sum; 4 NumArray(vector<int> &nums) { 5 sum.resize(nums.size(), 0); 6 sum[0] = nums[0]; 7 int len = nums.size(); 8 for(

【数组】Minimum Path Sum

题目: 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 any point in time. 思路: 设res[i][j]表示从左上角到grid[i][

HDU 5586 Sum

最大子串和 #include<cstdio> #include<cstring> const int maxn=100000+10; int n; int x[maxn]; int fx[maxn]; int a[maxn]; int sum[maxn]; int L[maxn],R[maxn]; const int INF=0x7FFFFFFF; int max(int a,int b) { if(a>b) return a; return b; } int main()

leetcode笔记：Combination Sum III

一. 题目描述 Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Ensure that numbers within the set are sorted in ascending order. E