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这题是在32. Search in Rotated Sorted Array I的基础上，处理特殊情况有重复数据的问题，有两个代表例子
如3,1,1和1,1,3,1只需在原基础上考虑这两种请求，完善上一题的分类思想，还有就是二分查找的迭代会完成接
下来的事情
class Solution {
public:
bool search(vector<int>& nums, int target) {
int low,high;
int media;
low = 0;
high = nums.size() - 1;
while(low <= high){
media = (high+low) / 2;
if(nums[media] == target)
return true;
if(nums[media] < nums[high])
if(nums[media] < target && target <= nums[high])
low = media + 1;
else
high = media - 1;
else if(nums[media] > nums[high])
if(nums[low] <= target && target < nums[media])
high = media - 1;
else
low = media + 1;
else
if(nums[high] != nums[low])
high = media - 1;
else
high = high - 1;
}
return false;
}
};
LeetCode:81. Search in Rotated Sorted Array II
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原文地址：http://www.cnblogs.com/jackes/p/5313951.html
