3Sum Closest & 3Sum Smaller

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.

Notice

You may assume that each input would have exactly one solution.

Example

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).

 public class Solution {
     /**
      * @param numbers: Give an array numbers of n integer
      * @param target : An integer
      * @return : return the sum of the three integers, the sum closest target.
      */
     public int threeSumClosest(int[] num ,int target) {
         int tempClosestSum = Integer.MAX_VALUE;
         int diff = Integer.MAX_VALUE;
         if(num == null || num.length < ) {
             return tempClosestSum;
         }
         Arrays.sort(num);
         for (int i = ; i < num.length - ; i++) {
             if (i !=  && num[i] == num[i - ]) {
                 continue; // to skip duplicate numbers; e.g [0,0,0,0]
             }

             int left = i + ;
             int right = num.length - ;
             while (left < right) {
                 int sum = num[left] + num[right] + num[i];
                 if (sum - target == ) {
                     return target;
                 } else if (sum - target < ) {
                     if (Math.abs(sum - target) < Math.abs(tempClosestSum - target)) {
                         tempClosestSum = sum;
                     }
                     left++;
                 } else {
                     if (Math.abs(sum - target) < Math.abs(tempClosestSum - target)) {
                         tempClosestSum = sum;
                     }
                     right--;
                 }
             }
         }
         return tempClosestSum;
     }
 }

3Sum Smaller

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]
[-2, 0, 3]

分析：https://segmentfault.com/a/1190000003794736
解题思路和3SUM一样，也是先对整个数组排序，然后一个外层循环确定第一个数，然后里面使用头尾指针left和right进行夹逼，得到三个数的和。如果这个和大于或者等于目标数，说明我们选的三个数有点大了，就把尾指针right向前一位（因为是排序的数组，所以向前肯定会变小）。
如果这个和小于目标数，那就有right - left个有效的结果。为什么呢？因为假设我们此时固定好外层的那个数，还有头指针left指向的数不变，那把尾指针向左移0位一直到左移到left之前一位，这些组合都是小于目标数的。

     public int threeSumSmaller(int[] nums, int target) {
         // 先将数组排序
         Arrays.sort(nums);
         int cnt = ;
         for (int i = ; i < nums.length - ; i++) {
             int left = i + , right = nums.length - ;
             while (left < right) {
                 int sum = nums[i] + nums[left] + nums[right];
                 // 如果三个数的和大于等于目标数，那将尾指针向左移
                 if (sum >= target) {
                     right--;
                     // 如果三个数的和小于目标数，那将头指针向右移
                 } else {
                     // right - left个组合都是小于目标数的
                     cnt += right - left;
                     left++;
                 }
             }
         }
         return cnt;
     }