# [LeetCode] 3Sum Smaller 三数之和较小值

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]


Could you solve it in O(n2) runtime?

// O(n^3)
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;
}
};

// 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;
}
};

3Sum Closest

3Sum

posted @ 2016-03-02 15:16 Grandyang 阅读(...) 评论(...) 编辑 收藏