[LeetCode 769] Max Chunks To Make Sorted_c++ max chunks to make sorted-优快云博客

给定一个数组，它是的排列，我们将数组分割成一些“块”并分别对每个块进行排序。连接它们后，结果等于排序后的数组。求我们最多可以将数组分成多少个块。本文分析解题思路并提供C++代码实现。

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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].

分析

这道题意思是将一个数组最多分成多少份，才会使得各个子数组排序后拼接成的最终数组是排好序的。

如果子数组排好序并且拼接后要求总的数组也是排好序的，那么就要求子数组数组arr[start, end]的值都在[start, end]之间，即

start<=arr[i]<=end, if start <= i <= end

那么下面我们就找最小[start, end]的范围能够保证这个关系。记录一个maxIndex = 0，遍历arr，当arr[i] > maxIndex，表明此时需要更新当前子数组的end范围到maxIndex，一直遍历到index == maxIndex，表明此时子数组的所有的arr[i] 都是小于maxIndex的。那么这个时候就找到了一个子数组的边界，res ++。

Code

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int maxIndex = 0;
        int res = 0;
        
        int len = arr.size();
        
        for (int i = 0; i < len; i ++)
        {
            if (arr[i] > maxIndex)
                maxIndex = arr[i];
            
            if (i == maxIndex)
                res ++;
        }
        
        return res;
    }
};

运行效率

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Max Chunks To Make Sorted.

Memory Usage: 8.4 MB, less than 57.60% of C++ online submissions for Max Chunks To Make Sorted.