259. 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]

Follow up:

Could you solve it in O(n2) runtime?

public class Solution {

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;

}

}