标签：isp turn none question ati dex tar integer audio
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?
解法1: 暴力brute force, 找出所有组合的和，然后找出小于target的组合，O(n^3)
解法2: 双指针， O(n^2)
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
if (nums.size() < 3) return 0;
int res = 0, n = nums.size();
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 2; ++i) {
int left = i + 1, right = n - 1;
while (left < right) {
if (nums[i] + nums[left] + nums[right] < target) {
res += right - left;
++left;
} else {
--right;
}
}
}
return res;
}
};
类似题目：
[LeetCode] 16. 3Sum Closest 最近三数之和
[LeetCode] 259. 3Sum Smaller 三数之和较小值
标签：isp turn none question ati dex tar integer audio
原文地址：https://www.cnblogs.com/lightwindy/p/9532678.html
