文章目录
  1. 1. 题意描述
  2. 2. 题意翻译
  3. 3. 代码参考

题意描述

This question is the same as “Max Chunks to Make Sorted” except the integers of the given array are not necessarily distinct, the input array could be up to length 2000, and the elements could be up to 10**8.

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]

题意翻译

这道题是Max Chunks To Make Sorted的翻版,但和之前不同的是这里的数组的元素是有重复的,方法和之前的差不多,但区别的一点是得有一个变量upper来表示当前数组元素是否满足条件,条件可以理解为不能大于已经出现的元素的最大值,这样可以解决元素重复的问题。

代码参考

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class Solution {
public int maxChunksToSorted(int[] arr) {
// the length of arr
int len = arr.length;
//
int[] index_max = new int[len];
// travel
int max = 0;
for (int i=0 ; i<len ; i++){
max = Math.max(max , arr[i]);
index_max[i] = max;
}

int[] sorted = arr.clone();
// sort
Arrays.sort(sorted);
// compare
int count = 0;
int upper = Integer.MAX_VALUE;

for (int i=len-1 ; i>=0 ; i--){
if (sorted[i] == index_max[i]){
if(sorted[i] > upper){
continue;
}

count++;
upper = arr[i];
}
}
return count;
}
}
文章目录
  1. 1. 题意描述
  2. 2. 题意翻译
  3. 3. 代码参考