LeetCode: Find Minimum in Rotated Sorted Array

Title:

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.

You may assume no duplicate exists in the array.

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0;
        int right = nums.size()-1;
        while (left < right && nums[left] > nums[right]){
            int mid = (left + right) / 2;
            if (nums[mid] < nums[left])
                right = mid;
            else 
                left = mid + 1;
        }
        return nums[left];
    }
};

II

The array may contain duplicates.

只需要修改while语句,并添加一个if语句。 [1,0,1,1,1]

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0;
        int right = nums.size()-1;
        while (left < right && nums[left] >= nums[right]){
            int mid = (left + right) / 2;
            if (nums[mid] < nums[left])
                right = mid;
            else if (nums[mid] > nums[left])
                left = mid+1;
            else 
                left += 1;
        }
		return nums[left];
    }
};

  

原文地址:https://www.cnblogs.com/yxzfscg/p/4530399.html