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| class Solution {
public:
int numberOfSubarrays(vector<int>& nums, int k) {
int n = nums.size();
if (n < k)
return 0;
int left = 0, right = 0, ans = 0;
int oddcount = nums[0] % 2 == 1 ? 1 : 0;
int leftcount = 0, rightcount = 0;
while (left < n) {
if (oddcount == k) {
while (right < n - 1 && nums[right + 1] % 2 == 0) {
right++;
rightcount++;
}
while (left <= right && left < n - 1 && nums[left] % 2 == 0) {
left++;
leftcount++;
}
ans += (leftcount + 1) * (rightcount + 1);
leftcount = 0;
rightcount = 0;
if (right < n - 1) {
right++;
oddcount++;
}
if (left <= right && left < n - 1) {
left++;
oddcount--;
} else if (left == n - 1)
break;
}
if (oddcount < k && right < n - 1) {
right++;
if (nums[right] % 2 == 1)
oddcount++;
} else if (oddcount < k && right >= n - 1)
break;
}
return ans;
}
};
|