16. 3Sum Closest

一,题目

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

方法一:枚举+剪枝

方法二:two pointer

方法二比方法一了一层循环

三,代码

public  int threeSumClosest(int[] nums, int target) {if (nums == null || nums.length < 3) {return 0;}int len = nums.length;int diff = Integer.MAX_VALUE;int sum = 0;Arrays.sort(nums);for (int i = 0; i < len - 2; i++) {for (int j = i + 1; j < len - 1; j++) {int right = j+1;while (right < len) {int tempSum = nums[i] + nums[j] + nums[right];int tempDiff = tempSum - target;int abs = Math.abs(tempDiff);if (abs < diff) {sum = tempSum;diff = abs;} if(tempDiff > 0 && tempDiff > diff){ break; }right++;}}}return sum;}public  int threeSumClosest2(int[] nums, int target){if (nums == null || nums.length < 3) {return 0;}Arrays.sort(nums);int len = nums.length;int bestSum = nums[0]+nums[1]+nums[2];for(int i=0; i<len-2 ;i++){int left = i+1;int right = len -1;while(left < right){int sum = nums[i]+nums[left]+nums[right];if(Math.abs(sum - target) < Math.abs(bestSum-target)){bestSum = sum;}if(sum > target){right --;}else{left++;}}}return bestSum;}