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.

 

由于是有序数组被旋转了，因此数组的性质也就发生了变化

要进行二分查找我们需要的数，进行如下步骤：

1. 取中间的数，比较 a[mid] 和 a[low]，若a[low] < a[mid]，说明旋转点在mid右边，则有a[low...mid]是有序数组，若target在a[low]与a[mid]之间，则在low和mid-1中继续找，否则在mid+1和high之间找

2. 若a[low] > a[mid]，说明旋转点在mid左边，那么a[mid...high]是有序数组，如果target在a[mid]和a[high]之间，则在mid+1和high之间继续找，否则在low和mid-1继续找

代码如下：

class Solution {
public:
    int search(int A[], int n, int target) {
        int low = 0, high = n-1;
        while( low <= high ) {
            int mid = low + (high-low) / 2;
            if( target == A[mid] ) return mid;
            if( A[low] < A[mid] ) { //若旋转点在右边，则A[low...mid]是有序数组
                if( A[low] <= target && target < A[mid] )
                    high = mid - 1;
                else
                    low = mid + 1;
            }
            else if( A[low] > A[mid] ) {    //若旋转点在左边，则A[mid...high]是有序数组
                if( A[mid] < target && target <= A[high] )
                    low = mid + 1;
                else
                    high = mid - 1;
            }
            else ++low; //low和mid指向同一个位置时
        }
        return -1;
    }
};

Search in Rotated Sorted Array II

 

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

 

直接修改上题代码。。。。AC......