LeetCode 3Sum Closest JS实现

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).

 //比较函数
 function sortNum(a,b)
 {
     return a-b;
 }
var threeSumClosest = function(nums, target) {
     switch (nums.length)  
        {  
        case 0:   
            return 0;  
        case 1:  
            return nums[0];  
        case 2:  
            return nums[0]+nums[1];  
        default:  
            break;  
        } 
    var k=nums.length-1;
    var arrNum=nums.sort(sortNum);   //先对数组排序
    var currentSum=arrNum[0]+arrNum[1]+arrNum[2];   //给定初始比较值
    var closeSum=currentSum;
    var diff =Math.abs(currentSum-target);  

        for (var i = 0; i < k-1; i++)    //先固定第一个数，和最后一个数，若三个数的和大于目标值，则k减1；若小于目标值加1；j和k值相等了，则i加1，进入外层下次循环，k值一定要重置；
        {  
           k=nums.length-1;     //k要重置
            for (var j = i+1; j < k;)  
            {   
                currentSum = arrNum[i] + arrNum[j] + arrNum[k];  
                if (currentSum === target)  
                {  
                    return currentSum;  
                }  
                if (Math.abs(currentSum-target) < diff)  
                {  
                    closeSum = currentSum;  
                    diff = Math.abs(currentSum - target);  
                }  
                if (currentSum < target)  
                {  
                    j++;  
                }  
                else 
                {  
                    k--;  
                }  
            }  
        }  
    return closeSum;
};