Leecode Week8: 3Sum Closest-优快云博客

本文介绍了一个中等难度的算法问题3SumClosest，旨在寻找整数数组中三个数的组合，使其总和最接近给定的目标值。通过排序和双指针技术实现了高效的解决方案。

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Week8: 3Sum Closest

Difficulty:Medium

1.Problem

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).

2.Algorithm

1.首先，对input进行排序

2.然后，遍历所有可能作为三个数之中的第一个数的元素。

3.对于每一个2步骤中的元素，对其后面的元素进行双向遍历以查找使得三个元素的和离target更近的元素，同时也进行了去除一些重复数字的工作

3.Solution

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int size = nums.size();
        sort(nums.begin(),nums.begin()+size);
        int result = nums[0] + nums[1] + nums[2];
        for(int i = 0; i < size - 2; i++) {
            if(i > 0 && nums[i] == nums[i-1]) {
                continue;
            }
            int start = i + 1, end = size - 1;
            while(start < end) {
                int sum = nums[i]+nums[start]+nums[end];
                if(abs(target-result) > abs(target-sum)) {
                    result = sum;
                    if(result == target) return result;
                    while(start < end && nums[start] == nums[start+1]) start++;
                    while(start < end && nums[end] == nums[end-1]) end--;
                }
                if(sum < target) start++;
                else end--;
            }
        }
        return result;
    }
};