259. 3Sum Smaller
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.
Example:
Input: nums = [-2,0,1,3], and target = 2
Output: 2
Explanation: 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?
- break three sum to two sum
public int threeSumSmaller(int[] nums, int target) {
Arrays.sort(nums);
int result = 0;
for(int i = 0; i < nums.length; i++) {
int t = target - nums[i];
int l = i+1, r = nums.length-1;
while(l < r) {
int sum = nums[l] + nums[r];
if(sum < t) {
result += r-l; // give up the smallest value
l++;
} else {
r--; // give up the largest value
}
}
}
return result;
}