题解 | #旋转数组的最小数字#

#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param nums int整型vector 
     * @return int整型
     */
    int minNumberInRotateArray(vector<int>& nums) {
        // write code here
        
        int high = nums.size() - 1;
        int low = 0;
        int middle = low;
        
        while (nums[low] >= nums[high]) {
            if (high - low == 1) {
                middle = high;
                break;
            }

            if (nums[low] == nums[high] && nums[high] == nums[middle]) {
                return MinInOrder(nums, low, high);
            }

            middle = (low + high) / 2;
            if (nums[middle] >= nums[low]) {
                low = middle;
            }
            if (nums[middle] <= nums[high]) {
                high = middle;
            }
            
        }

        return nums[middle];
    }

    int MinInOrder(vector<int> &nums, int low, int high) {
        int min = nums[low];
        for (int i = low + 1; i <= high; i++) {
            if (nums[i] < min) {
                min = nums[i];
            }
        }
        return min;
    }
};