leetCode 16.3Sum Closest (离给定值最近的三数之和) 解题思路和方法

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).

思路：这题和上一题3Sum一脉相承，只是变形了。所以此题整体解法上参考上题，细节上略作修改。

详细代码和注释如下。

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if(nums.length <= 2)
            return 0;

        int min = Integer.MAX_VALUE;//设初始值为最大
        int j,k,m,result = 0;
        Arrays.sort(nums);//排序
        for(int i = 0 ; i < nums.length - 1; i++){
            //如果数字和之前的数字相同，跳过
            if(i > 0 && nums[i] == nums[i-1]){
        		continue;
        	}
        	j = i+1;//初始值
        	k = nums.length -1;
        	while(j < k){
        	    m = nums[i] + nums[j] + nums[k];
        	    //更靠近target
        	    if(Math.abs(m - target) < min){
        	        min = Math.abs(m - target);//数据更新
        	        result = m;
        	        if(min == 0)//==0，找到目标值，没有比这更小的
        	        	return result;
        	        else{//如果没有达到0，则分情况标志位置变动
            	    	if(m - target < 0){//小于0左边+1
            	    		j++;
            	    		//消除重复的数字
                    	    while(j < k && nums[j] == nums[j-1]){
                    	        j++;
                    	    }
            	    	}
            	    	else{//大于0则右边位置减1
            	    		k--;
            	    		//消除重复值
            	    		while(j < k && nums[k] == nums[k+1]){
                    	        k--;
                    	    }
            	    	}
        	        }
        	    }//如果距离不比min更近，则直接分情况移动标记位置
        	    else{
        	    	if(m - target < 0)
        	    		j++;
        	    	else
        	    		k--;
    	        }
        	}
        }
        return result;
    }
}

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