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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.
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//思路首先：遍历一遍时间复杂度是O(N)
class Solution {
public:
int findMin(vector<int>& nums) {
int minum=nums[0];
for(int i=1;i<nums.size();i++)
{
if(nums[i]<minum)
minum=nums[i];
}
return minum;
}
};//思路首先：第二种方法还是直接遍历，
//按照题目所说数组应该是先上升，在下降，如果下降了，这个值就是最小值
class Solution {
public:
int findMin(vector<int>& nums) {
int minum=nums[0];
for(int i=1;i<nums.size();i++)
{
if(nums[i-1]>nums[i])
{
minum=nums[i];
break;
}
}
return minum;
}
};//思路首先：第三种方法，二分法
//
class Solution {
public:
int findMin(vector<int>& nums) {
if(nums.empty())
return 0;
if(nums.size() == 1)
return nums[0];
int low = 0, high = nums.size()-1;
while(low < high && nums[low] > nums[high])
{
int mid = low + (high-low)/2;
if(nums[mid] < nums[low])
high = mid;
else if(nums[mid] == nums[low])
return nums[high];
else
low = mid+1;
}
return nums[low];
}
};
<LeetCode OJ> Find Minimum in Rotated Sorted Array【153】
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原文地址：http://blog.csdn.net/ebowtang/article/details/50424986
