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.

Analyse: Binary search.

Runtime: 4ms.

 1 class Solution {
 2 public:
 3     int findMin(vector<int>& nums) {
 4         if(nums.empty()) return INT_MAX;
 5         
 6         int lowIndex = 0, highIndex = nums.size() - 1;
 7         while(lowIndex < highIndex) {
 8             int midIndex = lowIndex + (highIndex - lowIndex) / 2;
 9             
10             if(nums[midIndex] > nums[highIndex])
11                 lowIndex = midIndex + 1;
12             else
13                 highIndex = midIndex;
14         }
15         return nums[lowIndex];
16     }
17 };

分析：用二分查找法。mid = (low + high) / 2。考虑如下两种情形：

1. 如果nums[low] <= nums[mid]表示这一段区间是递增的，倘若target的值在两者之间，那么确定上界为mid - 1，否则下界为mid + 1；

2. 如果nums[low] > nums[mid]表示mid在最大值之后（举例： [4， 5， 6，0， 1， 2， 3]。数字0在6之后）如果target在nums[mid]和nums[high]之间，则可确定下界为mid + 1;否则上界为mid - 1。

运行时间7ms

 1 class Solution {
 2 public:
 3     int search(vector<int>& nums, int target) {
 4         if(nums.size() == 0) return -1;
 5         if(nums.size() == 1){
 6             if(nums[0] == target) return 0;
 7             else return -1;
 8         }
 9         
10         int low = 0, high = nums.size()-1;
11         while(low <= high){
12             int mid = (low + high) / 2;
13             if(nums[mid] == target) return mid;
14             if(nums[low] <= nums[mid]){
15                 if(nums[low] <= target && target < nums[mid]) high = mid;
16                 else low = mid + 1;
17             }
18             else{
19                 if(nums[mid] < target && target <= nums[high]) low = mid + 1;
20                 else high = mid;
21             }
22         }
23         return -1;
24     }
25 };