259 [LeetCode] 3Sum Smaller 三数之和较小值

题目：

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?

方法一：

class Solution {
public:int threeSumSmaller(vector<int>& nums, int target) {int res = 0;sort(nums.begin(), nums.end());for (int i = 0; i < int(nums.size() - 2); ++i) {int left = i + 1, right = nums.size() - 1, sum = target - nums[i];for (int j = left; j <= right; ++j) {for (int k = j + 1; k <= right; ++k) {if (nums[j] + nums[k] < sum) ++res;}}}return res;}
};

方法二：

双指针

class Solution2 {
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;}
};

转载于:https://www.cnblogs.com/250101249-sxy/p/10421748.html

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