LeetCode（153） Find Minimum in Rotated Sorted Array

题目

Total Accepted: 65121 Total Submissions: 190974 Difficulty: Medium

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.

分析

二分搜索的扩展应用。

寻找一个有序序列经旋转后的最小元素。

关键是掌握好，有序序列旋转后的元素之间的关系。

AC代码

class Solution {
public:
int findMin(vector<int>& nums) {
if (nums.empty())
return INT_MIN;

//元素个数
int size = nums.size();

int lhs = 0, rhs = size - 1;
while (lhs <= rhs)
{
//递增状态，返回低位值
if (nums[lhs] <= nums[rhs])
return nums[lhs];
//相邻，返回较小值
else if (rhs - lhs == 1)
return nums[lhs] < nums[rhs] ? nums[lhs] : nums[rhs];

//判断子序列
int mid = (lhs + rhs) / 2;
//右侧递增，则最小值位于左半部分
if (nums[mid] < nums[rhs])
{
rhs = mid;
}
//否则，最小值位于右半部分
else{
lhs = mid + 1;
}
}//while
}

//优化函数
int findMin2(vector<int>& nums)
{
if (nums.size() == 1)
return nums[0];

int lhs = 0, rhs = nums.size() - 1;

while (nums[lhs] > nums[rhs])
{
int mid = (lhs + rhs) / 2;
if (nums[mid] > nums[rhs])
lhs = mid + 1;
else
rhs = mid;
}//while
return nums[lhs];
}
};

GitHub测试程序源码