16. 3Sum Closest

寻找最接近目标的三数之和

最新推荐文章于 2019-05-28 15:42:21 发布

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本文介绍了一个算法问题：给定一个整数数组和一个目标值，找到数组中三个整数使得它们的和最接近目标值，并返回这个和。通过排序和双指针技巧实现了高效求解。

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).

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int res = target;
        std::sort(nums.begin(),nums.end());
        int ans = nums[0] + nums[1] + nums[2];
        for(int i = 0; i < nums.size(); ++i){
            int front = i+1, back = nums.size()-1;
            while(front < back){
                if(nums[i] + nums[front] + nums[back] < res){
                    if(res - (nums[i] + nums[front] + nums[back]) < abs(ans - res))
                    ans = (nums[i] + nums[front] + nums[back]);
                    front++;
                    
                }
                else if(nums[i] + nums[front] + nums[back] > res){
                    if((nums[i] + nums[front] + nums[back]) - res < abs(ans - res))
                    ans = (nums[i] + nums[front] + nums[back]);
                    back--;
                }
                else
                return res;
                
                while(front < back && nums[front] == nums[front+1])  // 错写成nums[front] == nums[front++].... sha
                front++;
                while(front < back && nums[back] == nums[back-1])
                back--;
            }
            
            while(i+1 < nums.size() && nums[i] == nums[i+1])
            ++i;
        }
        return ans;
        
    }
};