题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array/
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
).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
思路:由于是旋转数组,要确定应该往左还是往右搜索比较麻烦,我的解决方式是两个二分,一个二分找到右数组的左端点,这样就确定了两个数组的分界,然后根据target是不是比数组最后一个值大来确定要搜索左还是右数组。
如果target比数组最后一个值大,则说明target只能在左数组里搜索。
如果target比数组最后一个值小,则说明target要在右数组里搜索。
我觉得这样分开比较容易理解,也不容易出错。
代码如下:
class Solution { public: int search(vector<int>& nums, int target) { int len = nums.size(),left = 0, right = len -1; if(len == 0) return -1; while(left < right)//首先找到右子数组的左端点 { int mid = (left + right)/2; if(nums[mid] < nums[right]) right = mid; else left = mid + 1; } int low = (target > nums[len-1])?0:right;//新搜索数组的端点 int high = (target > nums[len-1])?(right-1):(len-1); while(low <= high)//然后判断搜索哪一个数组,是不是比较明了^@^ { int mid = (low + high)/2; if(nums[mid] < target) low = mid + 1; else high = mid - 1; } return nums[low]==target?low:-1; } };
而另外更为正统的做法是做一次二分,分析应该搜索左边还是右边。每次二分有三种情况:
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;
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: int search(vector<int>& nums, int target) { int len = nums.size(),left = 0, right = len -1; while(left <= right)//首先找到右子数组的左端点 { int mid = (left + right)/2; if(nums[mid] == target) return mid; if(nums[mid] < nums[right]) { if(target > nums[mid] && target <= nums[right]) left = mid + 1; else right = mid -1; } else { if(target >= nums[left] && target < nums[mid]) right = mid -1; else left = mid + 1; } } return -1; } };