LeetCode - 769. Max Chunks To Make Sorted

数组分块排序算法

转载于 2018-08-20 10:37:00 发布 · 92 阅读

本文介绍了一种针对特定数组（其元素为从0到数组长度减一的排列）的算法，该算法通过将数组分成若干个块（子数组），并对每个块进行单独排序，最终拼接成一个完全排序好的数组。探讨了如何确定能正确排序的最大块数，并提供了一个示例性的Java实现。

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.

class Solution {
    public int maxChunksToSorted(int[] arr) {
        if (arr == null)
            return 0;
        int max = 0, cnt = 0;
        for (int i=0; i<arr.length; i++) {
            max = Math.max(arr[i], max);
            if (max == i)
                cnt ++;
        }
        return cnt;
    }
}