153. Find Minimum in Rotated Sorted Array
153. Find Minimum in Rotated Sorted Array
难度：Medium
内容
题目链接：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array
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.
You may assume no duplicate exists in the array.
Example 1:
Input: [3,4,5,1,2]
Output: 1
Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0
思路
最简单的方法，挨个遍历，如果当前值小于前一个值，那么输出当前值即可，时间复杂度为O(n)。 二分查找法，时间复杂度为O(log(n))。
代码
快速简单的写了一个，提交，过了。
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];
}
};
二分查找的简单实现
class Solution {
public:
int findMin(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (nums[left] > nums[right]) {
int mid = (right + left) >> 1;
if (nums[mid] >= nums[left])
left = mid + 1;
else
right = mid;
}
return nums[left];
}
};
用迭代器来实现
class Solution {
public:
int findMin(vector<int>& nums) {
auto begin = nums.begin();
auto end = std::prev(nums.end());
while (*begin > *end) {
auto gap = (end - begin) >> 1;
*next(begin, gap) >= *begin ? advance(begin, gap + 1) : advance(end, -gap);
}
return *begin;
}
};
来源：https://github.com/xiaqunfeng/leetcode
