Lintcode: Find Minimum 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).

Find the minimum element.

Have you met this question in a real interview? 

Yes

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

Note

You may assume no duplicate exists in the array.

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Binary Search

Initial solution:
Time: O(n)
Space: O(1)
Compare adjacent elements one by one until find an element is less than the element before it.

class Solution {
public:
    /**
     * @param num: a rotated sorted array
     * @return: the minimum number in the array
     */
    int findMin(vector<int> &num) {
        // write your code here
        int len = num.size();
        if(len == 0) return 0;
        int pivot = 0;
        for(int i = 1; i < len; i++){
            if(num[i] < num[i - 1]) pivot = i;
        }
        return num[pivot];
    }
};

Binary search:
Set the num[0] as target to search, then start will point to the first element < num[0].
Note: when the sorted array is not rotated, no element < num[0], start will equal to the size of array.
We need to differentiate this situation.
Time: O(logn)
Space: O(1)

class Solution {
public:
    /**
     * @param num: a rotated sorted array
     * @return: the minimum number in the array
     */
    int findMin(vector<int> &num) {
        // write your code here
        int start = 0;
        if(num.size() == 0) return 0;
        int end = num.size() - 1;
        while(start <= end){
            int mid = (start + end) / 2;
            if(num[mid] < num[0]) end = mid - 1;
            if(num[mid] >= num[0]) start = mid + 1;
        }
        if(start == num.size()) return num[0];
        return num[start];
    }
};

Lintcode second:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

class Solution { public: /** * @param num: a rotated sorted array * @return: the minimum number in the array */ //8:25 -- 8:42 int findMin(vector<int> &num) { // write your code here if(num.size() == 0) return 0; int start = 0; int end = num.size() - 1; while(start <= end){ int mid = (start + end) / 2; //start 从左往右第一个<= num[0] //end 从右往左 第一个>num[0] if(num[mid] >= num[0]) start = mid + 1; else end = mid - 1; } if(start == num.size()) return num[0]; return num[start]; } };