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?
**解题思路 **
Binary Search Using n O(n2) runtime
注意： all hi in (lo, hi] will also satisfy the condition , 所以是 count += hi - lo 而不是count++;
public int threeSumSmaller(int[] nums, int target) {
int count = 0;
Arrays.sort(nums);
int len = nums.length;
for (int i = 0; i < len - 2; i++) {
int lo = i + 1, hi = len - 1;
while (lo < hi) {
if (nums[i] + nums[lo] + nums[hi] >= target) {
hi--;
} else {
count += hi - lo; // all hi in (lo, hi] will also satisfy the condition
lo++;
}
}
}
return count;
}
