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?
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.
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.
对数组分段排序后整合，求最多能分成几段。
解决：因为是permutation of [0, 1, ..., arr.length - 1]，那么有序的情况就是arr[i] == i。
本身有序的数可以自成一段；而无序的数找最大的那个错序数，作为分段的终点。
class Solution { public: int maxChunksToSorted(vector<int>& arr) { int res = 0; int i = 0; while (i<arr.size()) { if (arr[i] == i) { ++i; ++res; } else { int m = arr[i]; while (++i <= m) m = max(m, arr[i]); ++res; } } return res; } };
