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

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。然后依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。

详细代码例如以下：

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;
    }
}

兴许：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.

代码例如以下：

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;
    }
}