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【leetcode刷题笔记】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.


 

题解:还是按照二分的方法找target。

  1. 如果A[l] < A[mid],说明mid以左有序且都小于mid,如下图所示:这种情况下如果target在l和mid之间,那么需要把r重新置为mid;其他情况都需要到mid右端继续搜索。

     2.如果A[l] >= A[mid], 说明mid以右有序且都大于mid,如下图所示,如果target在mid和r之间,那么需要把l重新置为mid;其他情况都需要到mid左端继续搜索。

当l + 1 = r的时候,只要检查l和r所指向的元素是否等于target即可。

代码如下:

 1 public class Solution { 2     public int search(int[] A, int target) { 3         int l = 0; 4         int r = A.length - 1; 5          6         while(l + 1< r){ 7             int mid = l + (r-l)/2; 8             if(A[mid] == target) 9                 return mid;10             if(A[l]< A[mid] ){11                 if(A[mid] >= target && A[l] <= target)12                     r = mid;13                 else {14                     l = mid;15                 }16             }17             else {18                 if(target >= A[mid] && target <= A[r])19                     l = mid;20                 else {21                     r = mid;22                 }23             }24         }25         26         if(target == A[l])27             return l;28         if(target == A[r])29             return r;30         return -1;31     }32 }