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 < 3) {
            return tempClosestSum;
        }
        Arrays.sort(num);
        for (int i = 0; i < num.length - 2; i++) {
            // if (i != 0 && num[i] == num[i - 1]) {
            //     continue; // to skip duplicate numbers; e.g [0,0,0,0]
            // }

            int left = i + 1;
            int right = num.length - 1;
            while (left < right) {
                int sum = num[left] + num[right] + num[i];
                if (sum - target == 0) {
                    return target;
                } else if (sum - target < 0) {
                    left++;
                } else {
                    right--;
                }

                if (Math.abs(sum - target) < Math.abs(tempClosestSum - target)) {
                    tempClosestSum = sum;
                }
            }
        }
        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 = 0;
        for (int i = 0; i < nums.length - 2; i++) {
            int left = i + 1, right = nums.length - 1;
            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;
    }