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leetcode | 3Sum Closest

时间:2015-05-28 11:03:43      阅读:138      评论:0      收藏:0      [点我收藏+]

标签:array   sum   leetcode   

3Sum Closest : https://leetcode.com/problems/3sum-closest/

问题描述

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

解析

和 上篇 3Sum 思想相同,利用夹逼思想,枚举一个数nums[i],从j=i+1, k = nums.size()-1;夹逼,sum小于target:j++,sum大于target,k–
此问题不需考虑重复,因为只需要一个和的结果。

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int result = nums[0]+nums[1]+nums[2];//初始化
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size()-2; i++) {
            int j = i+1;
            int k = nums.size()-1;
            int sum;
            while (j < k) {
                sum = nums[i]+nums[j]+nums[k];
                if (abs(sum-target) < abs(result-target))
                    result = sum;
                if (sum < target)
                    j++;  //不需检测nums[j]==nums[j-1],可以重复,不需保存满足条件的元素
                else if (sum > target)
                    k--;
                else
                    return target;
            }
        //注:不可将第12、13行放在此处,因为此处的终止条件是j == k
        }
        return result;  
    }
};

leetcode | 3Sum Closest

标签:array   sum   leetcode   

原文地址:http://blog.csdn.net/quzhongxin/article/details/46117629

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