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:
    利用余数选择特殊位置元素
    CSS hack
    css选择器
    按yyyy-mm-dd格式输入一个日期,判断这是这一年的第几天
    输入不同year,month,打印月历
    java学习之多线程(二)
    java学习之多线程
    剑指offer--第一个只出现一次的字符
    剑指offer--两个链表的第一个公共结点
    剑指offer--最小的k个数
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11367286.html
Copyright © 2011-2022 走看看