zoukankan      html  css  js  c++  java
  • LeetCode 769. Max Chunks To Make Sorted

    原题链接在这里:https://leetcode.com/problems/max-chunks-to-make-sorted/

    题目:

    Given an array arr that is a permutation of [0, 1, ..., arr.length - 1], we split the array into some number of "chunks" (partitions), and individually sort each chunk.  After concatenating them, the result equals the sorted array.

    What is the most number of chunks we could have made?

    Example 1:

    Input: arr = [4,3,2,1,0]
    Output: 1
    Explanation:
    Splitting into two or more chunks will not return the required result.
    For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.
    

    Example 2:

    Input: arr = [1,0,2,3,4]
    Output: 4
    Explanation:
    We can split into two chunks, such as [1, 0], [2, 3, 4].
    However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
    

    Note:

    • arr will have length in range [1, 10].
    • arr[i] will be a permutation of [0, 1, ..., arr.length - 1].

    题解:

    From examples, we could tell if index moves to a point that all the value on its left including itself could cover all the indices, then it could be split into a chunk.

    [2, 0, 1]. when i = 0, all the values [2], index i = 0 is not covered by 0, then it is not a chunk.

    i = 1. values [2, 0], i = 1 is not covered by 1, then it is not a chunk.

    i = 2. values [2, 0, 1] all indices get covered. It could be split into a chunk.

    Then how to tell if all the indices are coverred. One way is to maintain the max, if max == i, then all indices are coverred. If max != i, then max must be larger than i. That means there is one value scanned larger, then there must be one index position not scanned yet.

    Time Complexity: O(n). n = arr.length. 

    Space: O(1).

    AC Java:

     1 class Solution {
     2     public int maxChunksToSorted(int[] arr) {
     3         int res = 0;
     4         if(arr == null || arr.length == 0){
     5             return res;
     6         }
     7         
     8         int max = 0;
     9         for(int i = 0; i<arr.length; i++){
    10             max = Math.max(max, arr[i]);
    11             if(max == i){
    12                 res++;
    13             }
    14         }
    15         
    16         return res;
    17     }
    18 }

    跟上Max Chunks To Make Sorted II.

  • 相关阅读:
    log日志----logging模块
    配置文件--configparser
    面向对象——进阶
    面向对象
    内置函数
    模块和包
    常用模块
    正则表达式
    递归函数
    内置函数+匿名函数
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11367286.html
Copyright © 2011-2022 走看看