原题描述:
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.
代码如下:
class Solution {
public:
int findm(vector<int> &num, int i, int j) {
if (i == j) return num[i];
if (i + 1 == j) return num[i] < num[j] ? num[i] : num[j];
int mid = (i+j) / 2;
if (num[i] < num[mid] && num[mid] < num[j]) return num[i];
if (num[i] < num[mid]) return findm(num, mid+1, j);
else return findm(num, i, mid);
}
int findMin(vector<int> &num) {
return findm(num, 0, num.size()-1);
}
};
备注:
1、递归的返回条件分两种情况,数组中剩余一个元素和两个元素,不考虑两个元素的情况容易错。比如例子:[2,1]
2、三目运算符? : 的运算级很低,在标准输出流是要用括号,return时不用
3、注意考虑边界条件