标签：style blog http io ar color os sp for
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
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.
The array may contain duplicates.
题意：求一个由不降序的数组转换成的新数组的最小值是多少
思路：第一种方法是：我们可以画出一个不降序的线段，然后有可能在任意点切掉，这个就是断点，所以可以看出出现第一个s[i]<s[i-1]的数的话，那么第i个数就是满足的
class Solution {
public:
int findMin(vector<int> &num) {
if (num.size() == 0)
return 0;
if (num.size() == 1)
return num[0];
for (int i = 1; i < num.size(); i++)
if (num[i] < num[i-1])
return num[i];
return num[0];
}
};第二种方法是：用vector自带的min_element暴力出来
class Solution {
public:
int findMin(vector<int> &num) {
if (num.size() == 0)
return 0;
vector<int>::iterator it = min_element(num.begin(), num.end());
return *it;
}
};
class Solution {
public:
int findMin(vector<int> &num) {
int left = 0;
int right = num.size() - 1;
while (left < right) {
if (num[left] < num[right])
break;
int mid = (left + right) / 2;
if (num[mid] > num[right])
left = mid + 1;
else if (num[left] > num[mid])
right = mid;
else
left = left + 1;
}
return num[left];
}
};
LeetCode Find Minimum in Rotated Sorted Array II
标签：style blog http io ar color os sp for
原文地址：http://blog.csdn.net/u011345136/article/details/41809679
