[Leetcode] 768. Max Chunks To Make Sorted II 解题报告

最新推荐文章于 2022-08-29 20:03:38 发布

原创最新推荐文章于 2022-08-29 20:03:38 发布 · 687 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Array

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了LeetCode第768题的解题报告，探讨如何将数组分割成多个部分并排序，以得到排序后的数组。通过分析，提出了一种在O(n)时间和O(n)空间复杂度内的解决方案，利用额外数组记录右侧最小值，从左到右扫描以确定最大可能的切块数。

题目：

Given an array arr of integers (not necessarily distinct), 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 = [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].

思路：

通过观察可知，如果在某个位置，左边最大的数比右边最小的数都大，那么说明就需要新分割出来一个chunk了。为了在线性时间复杂度内解决问题，我们额外定义一个数组right_min，其中right_min[i]表示在i的右边（包含i本身）的最小数。然后从左到右扫描数组，并记录截止当前的最大数。如果一旦发现其左边的最大数要大于等于右边的最小数，就新开始一个chunk（当然此时还需要更新left_max的值）。

算法的时间复杂度是O(n)，空间复杂度也是O(n)。

代码：

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        int n = arr.size();
        vector<int> right_min(n, 0);            // right_min[i] means the min value to its right
        right_min[n - 1] = arr[n - 1];
        for( int i = n - 2; i >= 0; --i) {
            right_min[i] = min(arr[i], right_min[i+1]);
        }
        int ans = 1, left_max = arr[0];
        for (int i = 1; i < n; ++i) {
            if (right_min[i] < left_max) {      // the right side has small value(s), hence cannot split
                left_max = max(left_max, arr[i]);
            }
            else {
                ++ans;
                left_max = arr[i];
            }
        }
        return ans;
    }
};