Add Date 2014-10-15
Find Minimum in Rotated Sorted Array
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.
这个很简单，因为没有重复数字，数组本质上还是有序的，用类似二分查找的方法复杂度O(log n)。记得考虑一下整个数组没有 rotated 的情况。
20号又加了有重复数字的题II，直接附II的 code 吧，II中需要考虑前中后三个数相等的情况，此时无法确定最小值在哪边，只能遍历一遍。
class Solution {
public:
int findMinN(vector<int> &num, int min, int max) {
int minN = num[min];
for(int i = min+; i <= max; ++i) {
if(num[i] < minN)
minN = num[i];
}
return minN;
}
int findMin(vector<int> &num, int min, int max) {
if(min == max || num[min] < num[max])
return num[min];
int mid = (max+min)>>;
if(num[mid] < num[min]) {
return findMin(num, min, mid);
}
else if(num[mid] > num[max]) {
return findMin(num, mid+, max);
}
else
return findMinN(num, min, max);
}
int findMin(vector<int> &num) {
int len = num.size();
return findMin(num, , len-);
}
};