1004. Max Consecutive Ones III
Given a binary array nums and an integer k, return the maximum number of consecutive 1’s in the array if you can flip at most k 0’s.
Example 1:
Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Example 2:
Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Constraints:
1 <= nums.length <= 105
nums[i] is either 0 or 1.
0 <= k <= nums.length
滑动窗口+前缀和
class Solution {
public:
int longestOnes(vector<int>& nums, int k) {
int n = nums.size();
vector<int> P(n + 1);
for (int i = 1; i <= n; ++i) {
P[i] = P[i - 1] + (1 - nums[i - 1]);
}
int ans = 0;
for (int right = 0; right < n; ++right) {
int left = lower_bound(P.begin(), P.end(), P[right + 1] - k) - P.begin();
ans = max(ans, right - left + 1);
}
return ans;
}
};