练习1 Shortest Unsorted Continuous Subarray

查找需要排序的最短子序列

题目:

给定一个整数数组，你需要找到一个连续的子阵。

你需要找到最短的子阵输出它的长度。

Example 1:

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

//思路：将原数组序列进行排序形成新的数组，将新数组和原数组进行比较，从左边开始对比，当两个数值不一致时，返回该值位置序号

//从右边开始比较，当元素值不一样时，返回该序列号，预期值=右边的序列号-左边序列+1 ，即中间的长度就是需要进行排序的元素，输出其长度。

My

public int findUnsortedSubarray(int[] nums) {

         int[] numsBySort = Arrays.copyOf(nums, nums.length);

         Arrays.sort(numsBySort);

         int left = 0;

         int right =0;

         int res = 0;

         //不需要排序的情况下

         if (Arrays.equals(nums, numsBySort)) {

                   return res;

         }

         for (int i = 0; i <nums.length; i++) {

                   if (nums[i]!=numsBySort[i]) {

                            if (i==nums.length-1) return nums.length;

                            left=i;

                            break;

                   }

         }

         for (int j = nums.length-1; j >0; j--) {

                   if (nums[j]!=numsBySort[j]) {

                            right=j;

                            break;

                   }

         }

         res=right-left+1;

         return res;

}

 

 

other https://discuss.leetcode.com/topic/89282/java-o-n-time-o-1-space

publicintfindUnsortedSubarray(int[] A) {

    int n = A.length, beg = -1, end = -2, min = A[n-1], max = A[0];

    for (int i=1;i<n;i++) {

      max = Math.max(max, A[i]);

      min = Math.min(min, A[n-1-i]);

      if (A[i] < max) end = i;

      if (A[n-1-i] > min) beg = n-1-i;

    }

    return end - beg +1;

}