题目要求: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.

 

代码如下:

class Solution {
public:
    int search(int A[], int n, int target) {
        
        int index = -1;

        //题目简化成找到数组A中的元素位置……
        for(int i = 0; i < n; i++){
            if(A[i] == target){
                index = i;
                break;
            }
        }
        
        return index;
    }
};

 

相关文章:

  • 2021-12-13
  • 2021-10-14
  • 2021-09-10
  • 2022-12-23
  • 2021-09-12
  • 2021-10-29
  • 2022-01-03
猜你喜欢
  • 2021-11-04
  • 2021-12-24
  • 2021-06-16
  • 2021-09-04
  • 2021-12-16
  • 2021-11-03
相关资源
相似解决方案