标签：exp cut values for number count nat public ges
Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s.
Example 1:
Input: A = [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: A = [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.
分析：
用两个指针: start, end. 不断往前移动end，当arr[end]是0的话，就增加count的个数。如果count的个数超过k，那么就往前移动start, 然后不断比较global_max和 end - start + 1.
1 class Solution { 2 public int longestOnes(int[] A, int K) { 3 int zeroCount = 0, start = 0, res = 0; 4 for (int end = 0; end < A.length; end++) { 5 if (A[end] == 0) { 6 zeroCount++; 7 } 8 while (zeroCount > K) { 9 if (A[start] == 0) { 10 zeroCount--; 11 } 12 start++; 13 } 14 res = Math.max(res, end - start + 1); 15 } 16 return res; 17 } 18 }
标签：exp cut values for number count nat public ges
原文地址：https://www.cnblogs.com/beiyeqingteng/p/12244468.html
