Suppose an array sorted in ascending order 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]).

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2] Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]Output: 0

Approach #1:

class Solution {public:    int findMin(vector
     
      & nums) {        int len = nums.size();        if (len == 1) return nums[0];        int l = 0, r = len-1;        while (l < r) {            int m = (l + r) / 2;            if (nums[m] < nums[r]) {                if (nums[m] > nums[l])                    r = m - 1;                else                     l = m + 1;            } else if (nums[m] > nums[l]) {                if (nums[m] < nums[r])                    r = m - 1;                else                     l = m + 1;            } else {                l++;            }        }        while (1) {            if (r > 0 && nums[r] > nums[r-1]) r--;            else return nums[r];        }    }};
     

Runtime: 

4 ms, faster than 47.61% of C++ online submissions for Find Minimum in Rotated Sorted Array.

 

 Optimization:

class Solution {public:    int findMin(vector
      
       & nums) {        int len = nums.size();        if (len == 1) return nums[0];        int l = 0, r = len-1;        while (l <= r) {            int m = (l + r) / 2;            if (nums[r] > nums[0])                return nums[0];            if (nums[m] > nums[m+1])                return nums[m+1];            if (nums[m] < nums[m-1])                return nums[m];            if (nums[m] > nums[0]) {                l = m + 1;            } else {                r = m - 1;            }        }        return -1;    }};
      

Runtime: 4 ms, faster than 47.61% of C++ online submissions for Find Minimum in Rotated Sorted Array.