769. Max Chunks To Make Sorted

标签：NPU   ssi   most   res   循环   tput   bsp   each   amp   

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].

解题思路：

一次循环，每次把循环中最大的数字找出来，如果在到达这个数字之前没有出现新数组的话，清零并且数量加1，重复操作。

1. class Solution {  
2. public:  
3.     int maxChunksToSorted(vector<int>& arr) {  
4.         if(arr.size()==0) return 0;  
5.         int max_partition=0;  
6.         int count=0;  
7.           
8.         for(int i=0;i<arr.size();i++){  
9.               
10.             if(arr[i]>max_partition) max_partition=arr[i];  
11.             if(max_partition == i) {count++;max_partition=0;}  
12.               
13.         }  
14.         return count;  
15.           
16.     }  
17. };  

769. Max Chunks To Make Sorted

标签：NPU   ssi   most   res   循环   tput   bsp   each   amp   

原文地址：https://www.cnblogs.com/liangyc/p/8819868.html