3Sum Smaller
Total Accepted: 8038 Total Submissions: 21165 Difficulty: Medium
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?
思路
public class Solution {
public int threeSumSmaller(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return 0;
}
int n = nums.length;
Arrays.sort(nums);
int count = 0;
for (int i = 0; i < n - 2; i++) {
int newTarget = target - nums[i];
int start = i + 1;
int end = n - 1;
while (start < end) {
if (nums[start] + nums[end ] < newTarget) {
count += end - start;
start++;
} else {
end--;
}
}
}
return count;
}
}