码迷,mamicode.com
首页 > 其他好文 > 详细

Leet Code OJ 1. Two Sum [Difficulty: Easy]

时间:2016-03-15 14:57:51      阅读:172      评论:0      收藏:0      [点我收藏+]

标签:

题目:
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.

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

翻译:
给定一个整形数组和一个整数target,返回2个元素的下标,它们满足相加的和为target。
你可以假定每个输入,都会恰好有一个满足条件的返回结果。

代码:

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] result=new int[2];
        for(int i=0;i<nums.length-1;i++){
            for(int j=i+1;j<nums.length;j++){
                if(nums[i]+nums[j]==target){
                    return new int[]{i,j};
                }
            }
        }
        return result;
    }
}

Leet Code OJ 1. Two Sum [Difficulty: Easy]

标签:

原文地址:http://blog.csdn.net/lnho2015/article/details/50889558

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!