【原创】leetCodeOj --- Find Minimum in Rotated Sorted Array II 解题报告

题目地址：

https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

题目内容：

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.

方法：

此题难点在于元素可重复，因此，关键在于如何分解出子问题。大概思路如下：

设首节点为C，中结点为A，末尾结点为D，则有：

C < D ? 返回 C

C == D ? 找到下一个为C不等的F，递归调用F,D。若不存在F，则返回C

C > D ? C < A ? ---> 递归调用A+1，D
C > A ? ---> 寻找A之前第一个和A不等的值B，若B > A则返回A，若B < A则递归调用C,B
C == A ? ---> 递归调用A+1，D

全部代码：

class Solution {
public:
int findMin(vector<int> &num) {
int len = num.size();
return trueStuff(num,0,len - 1);
}

int trueStuff(vector<int> &num,int start,int fin)
{
if (start == fin)
return num[start];
if (num[start] < num[fin])
return num[start];
if (num[start] == num[fin])
{
while (start < fin)
{
if (num[start] != num[fin])
break;
start ++;
}
if (start == fin)
return num[start];
else
return trueStuff(num,start,fin);
}
int mid = (start + fin) / 2;
if (num[start] < num[mid])
return trueStuff(num,mid + 1,fin);
if (num[start] > num[mid])
{
int tmp = mid;
while (num[tmp] == num[tmp-1])
tmp --;
tmp --;
if (num[tmp] > num[mid])
return num[mid];
else
return trueStuff(num,start,tmp);
}
if (num[start] == num[mid])
return trueStuff(num,mid + 1,fin);
}
};