1004. Max Consecutive Ones III

复制代码
package LeetCode_1004

/**
 * 1004. Max Consecutive Ones III
 * https://leetcode.com/problems/max-consecutive-ones-iii/description/
 *
 * 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.

Note:
1 <= A.length <= 20000
0 <= K <= A.length
A[i] is 0 or 1
 * */
class Solution {
    /*
    * solution: Sliding window, Time complexity:O(n), Space complexity:O(1)
    * */
    fun longestOnes(A: IntArray, K: Int): Int {
        var zeroCount = 0
        var left = 0
        var maxOneCount = 0
        for (right in A.indices) {
            if (A[right] == 0) {
                zeroCount++
            }
            while (zeroCount > K) {
                if (A[left] == 0) {
                    zeroCount--
                }
                left++
            }
            maxOneCount = Math.max(maxOneCount, right - left + 1)
        }
        return maxOneCount
    }
}
复制代码

 

posted @   johnny_zhao  阅读(96)  评论(0编辑  收藏  举报
编辑推荐:
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
阅读排行:
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布:重大改进与新特性概览!
· AI与.NET技术实操系列(二):开始使用ML.NET
· .NET10 - 预览版1新功能体验(一)
点击右上角即可分享
微信分享提示