581 Shortest Unsorted Continuous Subarray

最新推荐文章于 2019-03-31 16:27:06 发布

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最新推荐文章于 2019-03-31 16:27:06 发布

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分类专栏： LeetCode 文章标签： Array leetcode-java

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581 Shortest Unsorted Continuous Subarray

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 1:

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.

思路：两点法，遍历数组，从头开始寻找数组的最大值，从尾开始寻找数组的最小值，从前向后遍历过程中，如果当前值小于最大值，则更新end的位置，从后向前遍历过程中，如果当前值大于最小值则更新beg位置，最后返回end-beg+1。

代码

package Array;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int n = nums.length;
//        之所以设置beg=-1,end=-2是为了当前数组有序时，没有未排序的字数组时，可以直接返回end-beg+1=0
//        不需要做特殊处理
        int beg = -1;
        int end = -2;
        int max = nums[0];
        int min = nums[n - 1];
        for(int i=1;i<n;i++) {
            max = Math.max(nums[i], max);
            min = Math.min(nums[n - i - 1], min);
            if (max > nums[i]) {
                end = i;
            }
            if (min < nums[n - i - 1]) {
                beg = n - i - 1;
            }
        }
        return end - beg + 1;
    }
}