154. Find Minimum in Rotated Sorted Array II (旋转重复数组二分查找最小值)

Hard

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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.

The array may contain duplicates.

Example 1:

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

Example 2:

Input: [2,2,2,0,1]
Output: 0

1 class Solution {
 2 public:
 3     int findMin(vector<int>& nums) {
 4         int low = 0;
 5         int high = nums.size() - 1;
 6         while(low < high) {
 7             int mid = low + (high - low ) / 2;
 8             if(nums[mid] < nums[high]) {
 9                 high = mid;
10             } else if (nums[mid] > nums[high]){
11                 low = mid + 1;
12             } else { // nums[mid] = nums[high] ,并不能确定nums[mid] 究竟在最小值的左侧还是右侧，因此我们不能莽撞地忽略某一部分的元素.
13                 high--;
14             }
15         }
16         return nums[low];
17     }
18 };

 

class Solution {
public:
    int findMin(vector<int>& nums) {
        int low = 0;
        int high = nums.size()-1;
        while(low < high) {
            int mid = low + (high - low ) / 2;
            // 如果单调递增，如【1，2，3，4】 直接返回最小值。
            if (nums[low] < nums[high]) return nums[low];
            if(nums[low] > nums[mid]) {
                high = mid;
            } else if(nums[mid] > nums[low]) {
                low = mid+1;
            } else if (nums[mid] == nums[low]) {
                low+=1;
            }
        }
        return nums[low];
    }
};