You are given an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1].
We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.
Return the largest number of chunks we can make to sort the array.
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.
Constraints
- n == arr.length
- 1 <= n <= 10
- 0 <= arr[i] < n
- All the elements of arr are unique.
Solution
排序后 arr[i] == i ，每次遍历到的位置应该是前面块中的最大元素。
找每一块中最大的元素，如果最大的元素等于当前遍历的下标，找到该块，该元素及其前面的块排序都能到下标对应的位置。
int maxChunksToSorted(int* arr, int arrSize){
int i, max, chunks;
max = chunks = 0;
for (i = 0; i < arrSize; i++) {
if (arr[i] > max) max = arr[i];
if (max == i) {
chunks++;
max = 0;
}
}
return chunks;
}
