leetCode 33.Search in Rotated Sorted Array(排序旋转数组的查找) 解题思路和方法

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Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

思路：此题算法上不难，第一步计算旋转的步长。然后根据步长分情况得到升序的新的数组。然后二分查找target的索引，找到后再分情况返回元素的位置。

具体代码如下：

public class Solution {    public int search(int[] nums, int target) {        if(nums.length == 0){            return -1;        }                if(nums.length == 1){            if(nums[0] == target)                return 0;            return -1;        }                int rotate = 0;//旋转的步长        for(int i = nums.length - 1; i > 0; i--){            if(nums[i] < nums[i-1]){                rotate = i;                break;            }        }        int[] a = new int[nums.length];        //将数组按升序填充到新数组        for(int i = rotate; i < nums.length; i++){            a[i - rotate] = nums[i];//后面未旋转部分        }        for(int i = 0; i < rotate; i++){            a[i+nums.length-rotate] = nums[i];//前面旋转部分        }                int index = -1;        //二分查找        int low = 0;        int hight = nums.length - 1;        while(low <= hight){            int mid = (low + hight)/2;            if(a[mid] > target){                hight = mid - 1;            }else if(a[mid] == target){                index = mid;                break;            }else{                low = mid + 1;            }        }        if(index == -1)            return -1;        if(index + rotate > nums.length - 1)            return index + rotate - nums.length;        return index + rotate;    }}

后续：这一题实在解的有些多余了，最简单的就是一次遍历，找到返回index，找不到返回-1.

代码如下：

public class Solution {    public int search(int[] nums, int target) {        for(int i = 0; i < nums.length; i++){            if(nums[i] == target){                return i;            }        }        return -1;    }}

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