zoukankan      html  css  js  c++  java
  • LeetCode 487. Max Consecutive Ones II

    原题链接在这里:https://leetcode.com/problems/max-consecutive-ones-ii/

    题目:

    Given a binary array, find the maximum number of consecutive 1s in this array if you can flip at most one 0.

    Example 1:

    Input: [1,0,1,1,0]
    Output: 4
    Explanation: Flip the first zero will get the the maximum number of consecutive 1s.
        After flipping, the maximum number of consecutive 1s is 4.

    Note:

    • The input array will only contain 0 and 1.
    • The length of input array is a positive integer and will not exceed 10,000

    Follow up:
    What if the input numbers come in one by one as an infinite stream? In other words, you can't store all numbers coming from the stream as it's too large to hold in memory. Could you solve it efficiently?

    题解:

    Max Consecutive Ones进阶题目.

    用快慢指针. 每当nums[runner]是0时, zero count 加一.

    When zero count 大于可以flip最大数目时, move walker, 每当nums[walker]等于0, zero count 减一直到zero count 等于k.

    同时维护最大值res = Math.max(res, runner-walker).

    Time Complexity: O(nums.length). Space: O(1).

    AC Java:

     1 class Solution {
     2     public int findMaxConsecutiveOnes(int[] nums) {
     3         if(nums == null || nums.length == 0){
     4             return 0;
     5         }
     6         
     7         int res = 0;
     8         int count = 0;
     9         int walker = 0;
    10         int runner = 0;
    11         while(runner < nums.length){
    12             if(nums[runner++] != 1){
    13                 count++;
    14             }
    15             
    16             while(count > 1){
    17                 if(nums[walker++] != 1){
    18                     count--;
    19                 }
    20             }
    21             
    22             res = Math.max(res, runner-walker);
    23         }
    24         
    25         return res;
    26     }
    27 }

    Follow up说input是infinite stream, 不能把整个array放在memory中.

    When calculating res, can't move walker, since stream may be out of memory already.

    可以只用queue来记录等于0的index即可. 当queue.size() > k表示0的数目超过了可以flip的最大值,所以要dequeue.

    Time Complexity: O(n). Space: O(k).

    AC Java:

     1 class Solution {
     2     public int findMaxConsecutiveOnes(int[] nums) {
     3         if(nums == null || nums.length == 0){
     4             return 0;
     5         }
     6         
     7         int res = 0;
     8         int walker = 0;
     9         int runner = 0;
    10         LinkedList<Integer> que = new LinkedList<>();
    11         while(runner < nums.length){
    12             if(nums[runner++] != 1){
    13                 que.add(runner);
    14             }
    15             
    16             while(que.size() > 1){
    17                 walker = que.poll();
    18             }
    19             
    20             res = Math.max(res, runner-walker);
    21         }
    22         
    23         return res;
    24     }
    25 }

    类似Longest Substring with At Most Two Distinct CharactersMax Consecutive Ones III.

  • 相关阅读:
    全卷积网络(FCN)与图像分割
    Mac下编译tesseract报错 DotProductAVX can't be used on Android
    OSTU二值化算法
    ssh的用户配置文件config管理ssh会话
    SSD: Single Shot MultiBox Detector 编译方法总结
    论文笔记——A Deep Neural Network Compression Pipeline: Pruning, Quantization, Huffman Encoding
    LeetCode——Word Break
    C#多线程编程
    impinj 编程接口
    C# 委托实例(跨窗体操作控件)
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/6358630.html
Copyright © 2011-2022 走看看