[LeetCode] Search in Rotated Sorted Array II

https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

这道题目，需要注意有重复数字的情况，例如 1111 119 旋转为 119 1 111，那么[left,middle]区域将把9覆盖掉，这样，如果仍然使用之前的方法来做的话，如果target = 9，就直接把9漏掉了。同样，01111 11 旋转为 110 1 111，也会把0的查找漏掉。因此如果遇到nums[left]与nums[middle]相等的情况，我们需要对范围内的整个数组进行遍历来查找。代码如下：

// Author: Jian-xin Zhou

class Solution {
public:
bool search(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1, middle;

while (left + 1 < right) {
middle = (left + right) >> 1;

if (target == nums[middle]) {
return true;
}

// 两者相等时，如果范围取 [left,middle]，可能会有问题，例如 1119 1 1111 ，两个1之间夹了较大的9，二分可能会把9排除掉
if (nums[left] == nums[middle] ) {
for (int ix = left; ix <= right; ix++) {
if (target == nums[ix]) {
return true;
}
}

return false;
}

if (nums[left] < nums[middle]) {
if (nums[left] <= target && target < nums[middle]) {
right = middle;
} else {
left = middle;
}
}

if (nums[middle] <= nums[right]) {
if (nums[middle] < target && target <= nums[right]) {
left = middle;
} else {
right = middle;
}
}
}

if (target == nums[left]) {
return true;
}

if (target == nums[right]) {
return true;
}

return false;

}
};