【leetcode】Search in Rotated Sorted Array

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.

 

 

利用二分法，

如果left<A[mid]，说明从左边left开始到mid是有序的

 

如果left>A[mid]，说明右边从mid开始，到right是有序的

 

如果left=A[mid]，则说明left和mid在同一个位置，如果A[mid]!=target则下一步中left=left+1;

 

 class Solution {
 public:
     int search(int A[], int n, int target) {

         int left=;
         int right=n-;
         int mid;
         while(left<=right)
         {
             mid=(left+right)/;

             if(A[mid]==target) return mid;

             if(A[left]<A[mid])//left
             {
                 if(A[left]<=target&&target<A[mid])
                 {
                     right=mid-;
                 }
                 else
                 {
                     left=mid+;
                 }
             }
             else if(A[left]>A[mid])//right
             {
                 if(A[mid]<target&&target<=A[right])
                 {
                     left=mid+;
                 }
                 else
                 {
                     right=mid-;
                 }
             }
             else
             {
                 //最后一个判断语句可以一起放到第一个语句里，if(A[left]<=A[mid])
                 left=mid+;
             }

         }
         return -;
     }
 };