[LeetCode 解题报告]033. Search in Rotated Sorted Array
作者:互联网
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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1
考察:二分查找,首先我们数组nums=[0,1,2,4,5,6,7]的所有旋转情况,如下:
- 0,1, 2,4,5,6,7
- 1,2,4,5,6,7,0
- 2,4,5,6,7,0,1
- 4,5,6,7,0,1,2
- 5,6,7,0,1,2,4
- 6,7,0,1,2,4,5
- 7,0,1,2,4,5,6
可以发现,无论怎样的旋转,要么前半段有序,要么后半段有序;那么怎么判断是前半段有序还是后半段有序呢。
- nums[mid] < nums[right] 后半段有序
- nums[left] < nums[mid] 前半段有序
class Solution {
public:
int search(vector<int>& nums, int target) {
if (nums.size() < 1 || (nums.size() == 1 && nums[0] != target))
return -1;
int left = 0, right = nums.size()-1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
else if (nums[mid] < nums[right]) { //后半段有序
if (nums[mid] < target && nums[right] >= target)
left = mid + 1;
else
right = mid - 1;
}
else { // 前半段有序
if (nums[left] <= target && target < nums[mid])
right = mid - 1;
else
left = mid + 1;
}
}
return -1;
}
};
完,
标签:Search,target,nums,Rotated,mid,前半段,有序,Sorted,left 来源: https://blog.csdn.net/caicaiatnbu/article/details/102772077