[leetcode] 81. Search in Rotated Sorted Array II 解题报告

题目链接：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.

思路：和上题思路差不多，分析应该搜索左边还是右边。每次二分有四种情况：

1. nums[mid] = target，则可以返回mid

2. nums[mid] < nums[right]，说明在[mid, right]区间是右边递增的区间，然后判断target是否在这个区间内

1）如果nums[mid] < target <= nums[right]，说明target在右边区间里，则left = mid + 1;

2）否则在左边区间里，搜索左边区间，right = mid - 1;

3. nums[mid] > nums[right]，说明[elft, mid]区间是在左边的递增区间，然后判断target是否在这个左边区间里

1）如果nums[left] <= target < nums[mid]，说明target在这个区间里，则使right = mid - 1;

2）否则说明target在[mid, right]的不规则区间里，搜索右边区间，则使left = mid + 1;

4. nums[mid] == nums[right]，这种情况并不知道搜索哪一边，因为nums[right] != target，此时可以抛弃nums[right]，搜索[left, right-1]的区域

然后考虑边界条件：

如果数组长度为1，则left = right，mid = left=right，如果nums[mid] = target，则直接返回，否则会进入条件3，然后循环结束，返回-1；

代码如下：

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int len = nums.size(), left = 0, right = len-1;
        if(len == 0) return false;
        while(left <= right)
        {
            int mid = (left + right)/2;
            if(target == nums[mid]) return true;
            if(nums[mid] < nums[right])//
            {
                if(target > nums[mid] && target <= nums[right])//搜索右边有序数组
                    left = mid + 1;
                else
                    right = mid -1;
            }
            else if(nums[mid] > nums[right])
            {
                if(target >= nums[left] && target < nums[mid])//搜索左边有序数组
                    right = mid -1;
                else
                    left = mid + 1;
            }
            else
                right--;
        }
        return false;
    }
};

参考：http://bangbingsyb.blogspot.com/2014/11/leetcode-search-in-rotated-sorted-array.html