【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=0;
        int right=n-1;
        int mid;
        while(left<=right)
        {
            mid=(left+right)/2;
           
            if(A[mid]==target) return mid;
           
            if(A[left]<A[mid])//left
            {
                if(A[left]<=target&&target<A[mid])
                {
                    right=mid-1;
                }
                else
                {
                    left=mid+1;
                }
            }
            else if(A[left]>A[mid])//right
            {
                if(A[mid]<target&&target<=A[right])
                {
                    left=mid+1;
                }
                else
                {
                    right=mid-1;
                }
            }
            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+1;
            }
           
        }
        return -1;
    }
};