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?
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這種題要麼用map要麼幾個pointer。 有一個很巧妙的O(n^2)的算法: after sorting, if(i,j,k) is a valid triple, then i, j-1, k … i, i+1, k is also valid. there is no need to count again.
比如:sort之後的vector爲:-2, 0, 1, 4, 5, 7, 9 target = 5
if i = 0, j = 1, k = 5 valid, which means when –k to 2 all combinations are valid, there are total k-j combinations. All we have to do is increasing j.
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
int n = nums.size(), cnt = 0;
sort(nums.begin(), nums.end());
for(int i = 0; i<n-2; ++i){
if(nums[i] + nums[i+1] + nums[i+2] >= target) break;
int j = i+1, k = n-1;
while(j<k){
while(j<k && nums[i] + nums[k] + nums[j] >= target) --k;
cnt += k-j;
++j;
}
}
return cnt;
}
};