【LeetCode】581.Shortest Unsorted Continuous Subarray(easy)解题报告

【LeetCode】581.Shortest Unsorted Continuous Subarray(easy)解题报告

题目地址：https://leetcode.com/problems/shortest-unsorted-continuous-subarray/description/

题目描述：

  Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

  You need to find the shortest such subarray and output its length.

Example:

Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Note：

1. Then length of the input array is in range [1, 10,000].
2. The input array may contain duplicates, so ascending order here means <=.

题意：数组题，求不是正确顺序的中间部分数组长度。很多时候会考虑排序什么，解法一先排序，在比较，分别找到值不同的首尾index，得到长度。第二种解法借助单调栈。

Solution1:

public class Solution{
public int findUnsortedSubarray(int[] nums){
int[] snums = nums.clone();
Arrays.sort(snums);
int start = snums.length, end=0;
for(int i=0 ; i<snums.length ; i++）{
if(snums[i]!=nums[i]){
start = Math.min(start , i);
end = Math.max(end,i);
}
}
return (end - start>=0 ? end-start+1 ： 0);
}
}

Solution2:

/*
monotonic stack 单调栈
2 6 4 8 10 9 15

leftBound = Math.min(monotonic stack ops)
rightBound
shorts = rightBound - leftBound + 1
*/
class Solution {
public int findUnsortedSubarray(int[] nums) {
if(nums.length<2){
return 0;
}
Stack<Integer> stack = new Stack<>();
int leftBound = nums.length-1, rightBound = 0;
for(int i=0 ; i<nums.length ; i++){
if(stack.isEmpty() || nums[i]>= nums[stack.peek()]){
stack.push(i);
}else{
leftBound = Math.min(leftBound,stack.pop());
i--;
}
}
stack.clear();
for(int i=nums.length-1;i>=0;i--){
if(stack.isEmpty() || nums[i] <= nums[stack.peek()]){
stack.push(i);
}else{
rightBound = Math.max(rightBound , stack.pop());
i++;
}
}
return rightBound - leftBound > 0 ? rightBound - leftBound + 1 : 0;
}
}

Date：2017年11月7日