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

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;
    }
}

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

标签：leetcode   二分查找   search   

原文地址：http://blog.csdn.net/xygy8860/article/details/46806349