LeetCode 33. Search in Rotated Sorted Array

LeetCode 33. Search in Rotated Sorted Array

Description:

Suppose an array sorted in ascending order 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.

分析：

这道题当然是有时间复杂度限制的，要不然直接一个for循环就搞定了。

所以我们考虑二分搜索算法。这道题也不是纯粹的二分搜索，题意的阐述说明我们要对常规的二分搜索做一点点修改。题意说明该数组是以某个位置为主，左右两边都是有序数组，且都为递增的。而二分搜索要求数组为按从小到大顺序的排序数组，所以我们分左右两边情况讨论：

首先，我们比较数组最左边的数和数组中间数的大小，即

nums[left] <= nums[mid]

若成立，说明中间数落在左边递增区间里，例如数组[4, 5, 6, 7, 0, 1, 2]，中间数为7，落在左边递增区间，接下来只需比较target是否大于等于nums[left]并且小于nums[mid]即可。若成立，则

right = mid - 1

；反之

left = mid + 1

.

反之，说明中间数落在右边递增区间里，例如数组[4, 5, 6, 0, 1, 2, 3]，中间数为0，落在右边递增区间，接下来比较target是否大于nums[mid]并且小于等于nums[right]即可。若成立，则

left = mid + 1

；反之

right = mid - 1

.

此过程不断循环，循环条件为

left <= right

，并且比较nums[mid]是否等于target，若等于，则返回mid，即为结果。

代码如下：

class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) return mid;
if (nums[left] <= nums[mid]) {
if (target >= nums[left] && target < nums[mid])
right = mid - 1;
else left = mid + 1;
}
else {
if (target > nums[mid] && target <= nums[right])
left = mid + 1;
else right = mid - 1;
}
}
return -1;
}
};

Another Solution: Discuss上的一个解法

class Solution {
public:
int search(vector<int>& nums, int target) {
int n = nums.size();
int left = 0, right = n - 1;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] > nums[right]) left = mid + 1;
else right = mid;
}
int small = left;
left = 0;
right = n - 1;
while (left <= right) {
int mid = (left + right) / 2;
int realmid = (mid + small) % n;
if (nums[realmid] == target) return realmid;
if (nums[realmid] < target) left = mid + 1;
else right = mid - 1;
}
return -1;
}
};