No.016 3Sum Closest

16. 3Sum Closest

• Total Accepted: 86565
• Total Submissions: 291260
• Difficulty: Medium

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：

　　本题的解法和上一次3Sum的解法有点类似，我们可以先将数组排序，然后将数组中的数从左到右依次确定为第一个数，在其右边的数中寻找最接近的target的数即可。重要的是判断逻辑的思路。代码如下：

 public int threeSumClosest(int[] nums, int target) {
     if (nums == null || nums.length < 3) {
         return 0;
     }

     Arrays.sort(nums);

     int closest = nums[0]+nums[1]+nums[2] ;

     for(int i = 0 ; i < nums.length-2 ; i++){
         int l = i+1 ;
         int r = nums.length-1 ;
         while(l < r){
             int sum = nums[i] + nums[l] + nums[r] ;
             if(sum == target){
                 return sum ;
             }else if(sum < target){
                 /*
                  * 三种情况：
                  *     1、closest < sum < target --->closest = sum
                  *     2、sum < target < closest --->| target-sum < closest-target ---> closest = sum
                  *                                   | target-sum >= closest-target --> closest不变
                  *     3、sum <= closest <= target ---> closest不变
                  */
                 if(sum > closest){
                     //情况1
                     closest = sum ;
                 }else if(closest > target){
                     //情况2
                     if(target-sum < closest-target){
                         //情况2.1，
                         closest = sum ;
                     }
                     //情况2.2不用变化
                 }
                 //情况3不同变化
                 l++ ;
             }else{
                 //分析同上
                 if(sum < closest){
                     closest = sum ;
                 }else if(closest < target){
                     if(sum-target <= target-closest){
                         closest = sum ;
                     }
                 }
                 r-- ;
             }
         }

     }
     return closest ;
 }