153. Find Minimum in Rotated Sorted Array

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<int>& 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<int>& 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.