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154. Find Minimum in Rotated Sorted Array II

154. Find Minimum in Rotated Sorted Array II

难度:Hard

内容

题目链接:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii

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

Note:

思路

这题竟然是hard,好尴尬。最简单时间复杂度为O(n)的方法,我把前一题的代码一行不动的挪过来,Accepted。 二分查找法,时间复杂度为O(log(n))。

代码

O(n),和前一题代码一样

class Solution {
public:
    int findMin(vector<int>& nums) {
        int res_idx = 0;
        for (int i = 1; i < nums.size(); ++i)
            if (nums[i] < nums[i-1]) {
                res_idx = i;
                break;
            }
        return nums[res_idx];
    }
};

二分法实现

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0, right = nums.size() - 1;

        while (left < right && nums[left] >= nums[right]) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == nums[left]) {
                ++left;
            } else if (nums[mid] < nums[left]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return nums[left];
    }
};

迭代器二分查找的实现

class Solution {
public:
    int findMin(vector<int>& nums) {
        auto left = nums.begin();
        auto right = std::prev(nums.end());
        while (*left >= *right && left < right) {
            auto gap = (right - left) >> 1;
            if (*next(left, gap) == *left)
                advance(left, 1);
            else if (*next(left, gap) < *left)
                advance(right, -gap);
            else
                advance(left, gap + 1);
        }
        return *left;
};

来源:https://github.com/xiaqunfeng/leetcode



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