768. Max Chunks To Make Sorted II

问题：

给一个数组（这一题可以有重复的数字），我们将其最多分成几块，各自进行全排序，能得到整体有序的数列？

数值大小范围增大，元素个数增多

Example 1:
Input: arr = [5,4,3,2,1]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [5, 4], [3, 2, 1] will result in [4, 5, 1, 2, 3], which isn't sorted.


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


Note:
arr will have length in range [1, 2000].
arr[i] will be an integer in range [0, 10**8].

　　

解法：

构建两个数组：

* 左边最大值数组 leftmax

* 右边最小值数组 rightmin

左边最大【i】<=右边最小【i+1】(因为存在可重复数字)【满足有序的情况】

那么res++

代码参考：

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int n=arr.size(), res=1;
        vector<int> rightmin = arr, leftmax = arr;
        for(int i=1; i<n; i++){
            leftmax[i]=max(leftmax[i],leftmax[i-1]);
        }
        for(int i=n-2; i>=0; i--){
            rightmin[i]=min(rightmin[i],rightmin[i+1]);
        }
        for(int i=0; i<n-1; i++){
            if(rightmin[i+1]>=leftmax[i]) res++;
        }
        
        return res;
    }
};

可简化构建leftmax的过程，合并至最后一步的判断大小中，且简化数组为单个当前最大值 curmax

代码参考：

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int n=arr.size(), res=1;
        vector<int> rightmin = arr;
        int curmax=arr[0];
        for(int i=n-2; i>=0; i--){
            rightmin[i]=min(rightmin[i],rightmin[i+1]);
        }
        for(int i=0; i<n-1; i++){
            curmax=max(curmax, arr[i]);
            if(rightmin[i+1]>=curmax) res++;
        }
        
        return res;
    }
};