16. 3Sum Closest

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).

在给定数列中找出三个数，使和最接近 target

还是先排序，再左右夹逼

public class Solution {  
  
    public int threeSumClosest(int[] num, int target) {  
        Arrays.sort(num);  
        int ret = num[0] + num[1] + num[2];  
        int len = num.length;  
        for (int i = 0; i <= len - 3; i++) {  
            // first number : num[i]  
            int j = i + 1;  // second number  
            int k = len - 1;    // third number  
            while (j < k) {  
                int sum = num[i] + num[j] + num[k];  
                if (Math.abs(sum - target) < Math.abs(ret - target))  
                    ret = sum;  
                if (sum < target) {  
                    ++j;  
                } else if (sum > target) {  
                    --k;  
                } else {  
                    ++j;  
                    --k;  
                }  
            }  
        }  
        return ret;  
  
    }  
}  

题意：数组中每三个元素进行求和，找出所有和中大小最接近target的和，并返回这个和与target之间的差值。

解题思路：使用一个变量mindiff来监测和与target之间的差值，如果差值为0，直接返回sum值。

代码：

class Solution:
    # @return an integer
    def threeSumClosest(self, num, target):
        num.sort()
        mindiff=100000
        res=0
        for i in range(len(num)):
            left=i+1; right=len(num)-1
            while left<right:
                sum=num[i]+num[left]+num[right]
                diff=abs(sum-target)
                if diff<mindiff: mindiff=diff; res=sum
                if sum==target: return sum
                elif sum<target: left+=1
                else: right-=1
        return res