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;
      }
  };

原文地址：https://www.cnblogs.com/250101249-sxy/p/10421748.html