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).

看清楚是计算三个整形数的和返回，不是求最小差值。题目思路和这篇博文说到的一样，只是编码进行适当小的改变。http://blog.youkuaiyun.com/huruzun/article/details/38710261

package Three_Sum;

import java.util.Arrays;

public class Solution3 {
    public int threeSumClosest(int[] num, int target) {
        int diff = Integer.MAX_VALUE;
        int temp ;
        int flag = 1;
        Arrays.sort(num);
        for(int i=0;i<num.length-2;i++){
        	temp = twoSumDiff(num, target-num[i], i+1);
        	if(temp ==0){
        		return target;
        	}
        	//diff 是三个数和与目标的差值不带符号。
        	if(Math.abs(temp)<diff){
        		diff = Math.abs(temp);
        		if(temp<0)
        			flag = -1;
        		else {
					flag = 1;
				}
        	}
        }
        // 目标加上差值就为三个最接近数字的和
        return (diff*flag)+target;
    }
    // 返回两个值与给定目标的差值，这个值是由正负之分的。
    public int twoSumDiff(int[]num,int target,int start){
    	 int res = Integer.MAX_VALUE;
    	 int head = start;
    	 int tail = num.length-1;
    	 int flag = 1;
    	 while(head<tail){
    		 int temp = num[head]+num[tail];
    		 if(temp>target){
    			 if(res > Math.abs(temp-target)){
    				 res = Math.abs(temp-target);
    				 flag = 1;
    			 }
    			 tail--;
    		 }else if(temp < target){
    			 if(res > Math.abs(temp-target)){
    				 res = Math.abs(temp-target);
    				 flag = -1;
    			 }
    			 head++;
    		 }
    		 else {
				return 0;
			}
    	 }
    	 return res*flag;
    }
	public static void main(String[] args) {
		int []num = {-1 ,2, 1, -4};
		Solution3 s3 = new Solution3();
		System.out.println(s3.threeSumClosest(num, 1));
	}
}

代码中包含了main函数测试代码。