leetcode--16. 3Sum Closest

一、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).

二、思路分析

类似与之前的3sum问题：

先对S进行排序，建立三个索引 i,  j,  k;其中 i从0开始到size-2，j时刻比i大1，k从size-1开始，计算是sum = s[i]+s[j]+s[k],其中要消除重复，temp = sum -target;

循环i从0到size-2；

while循环直到j >=  k;

每个循环有三种情况：

如果temp = 0, 则sum == target此时最近为target本身；

如果temp < 0, 则j++,  不断靠近；

如果temp > 0, 则k--，不断靠近；

三、代码

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
    	int n = nums.size();
        int res;
    	int min = 520;//记录最近距离
        int temp;
        for (int i = 0; i < n; i++) {
        	for (int j = 0; j < n-i-1; j++) {
        		if (nums[j] > nums[j+1]) {
        			temp = nums[j];
        			nums[j] = nums[j+1];
        			nums[j+1] = temp;
        		}
        	}
        }

        int j = 0, k = 0;
        for (int i = 0; i < n-2; ++i) {
        	if (i > 0&&nums[i] == nums[i-1]) {
        		continue;
        	}
        	j = i+1;
        	k = n-1;
        	while(j < k) {
        		int sum = nums[i]+nums[j]+nums[k];
        		int temp = sum-target;
        		int tt = abs(temp);
		    	if (temp == 0) {
				return target;
		    	} else if(temp < 0) {
		    		if (min > tt) {
		    			min = tt;
                        		res = sum;
		    		}
		    		j++;
		    	} else {
		    		if (min > tt) {
		    			min = tt;
                        		res = sum;
		    		}
		    		k--;
		    	}
		    }
        }
        return res;
    }
};