LeetCode 3Sum Closest JS實現

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).

 //比較函數 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);   //先對數組排序    var currentSum=arrNum[0]+arrNum[1]+arrNum[2];   //給定初始比較值    var closeSum=currentSum;    var diff =Math.abs(currentSum-target);          for (var i = 0; i < k-1; i++)    //先固定第一個數，和最後一個數，若三個數的和大於目標值，則k減1；若小於目標值加1；j和k值相等了，則i加1，進入外層下次迴圈，k值一定要重設；        {             k=nums.length-1;     //k要重設            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;};