LeetCode:3Sum Closest

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3Sum Closest

Total Accepted: 80994 Total Submissions: 275223 Difficulty: Medium

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).

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思路：

暴力解法是三个循环i,j,k，判断（nums[i] + nums[i] + nums[k]）与target的最接近值，时间复杂度O(n^3)；

但是其实有许多重复的计算，就是在三个指针交叉指向的时候；

如：（i = 0,j=2,k=4）与（i=2,j=4,k=2）这些都是不必要的。

三个指针，first loop(0~n-1)，每次seond(从first+1开始)和third(从n-1开始)相互逼近，这样可以去掉重复值；

时间复杂度可降到O(n^2)。

java code:

public class Solution {    public int threeSumClosest(int[] nums, int target) {                int n = nums.length;        int closetSum = nums[0] + nums[1] + nums[n - 1];        Arrays.sort(nums);        for(int first = 0; first < n-2; first++) { //不同有两个指针指向同一位置，因此是n-2            //if(first > 0 && nums[first] == nums[first-1]) continue;//去重用，这里不需要            int second = first + 1, third = n - 1;            while(second < third) {                int curSum = nums[first] + nums[second] + nums[third];                if(curSum==target) return target;                if(Math.abs(target - curSum) < Math.abs(target - closetSum))                    closetSum = curSum;                if(curSum < target)                    second++;                else                    third--;            }        }        return closetSum;    }}

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