There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
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
Example 3:
Input: nums = [1], target = 0
Output: -1
Constraints:
1 <= nums.length <= 5000-$10^4$ <= nums[i] <= $10^4$- All values of
numsare unique. numsis an ascending array that is possibly rotated.$10^4$ <= target <= $10^4$
分析: 这道题采用分治法. 由于经过了反转, nums不是有序的, 因此首先要找到旋转中心.
int findRotIndex(vector<int> const &nums) {
int left = 0;
int right = nums.size()-1;
while (left <= right) {
int mid = left + (right-left)/2;
int rightVal = nums[mid+1];
if (rightVal < nums[mid]) {
return mid;
}
// 10 11 12 0 1 2 3
if (nums[mid] > nums[left]) {
left = mid+1;
} else {
right = mid;
}
}
return -1;
}
接下来可以判断target旋转中心左边还是右边序列, 采用二分查找进行搜索
int binarySearch(vector<int> const &nums, int left, int right, int const target) {
while (left <= right) {
int mid = left + (right-left)/2;
if (nums[mid] == target) return mid;
if (target > nums[mid]) {
left = mid+1;
} else {
right = mid-1;
}
}
return -1;
}
本题的答案:
class Solution {
public:
int search(vector<int>& nums, int target) {
if (nums.front() <= nums.back()) {
// strictly increasing order
return binarySearch(nums, 0, nums.size()-1, target);
}
int const rotIdx = findRotIndex(nums);
// std::cout << "rot idx: " << rotIdx << " val: " << nums[rotIdx] << std::endl;
if (target == nums[rotIdx]) return rotIdx;
if (target < nums.front()) { // not a comparison against the value at the rotation index!
return binarySearch(nums, rotIdx+1, nums.size()-1, target);
} else {
return binarySearch(nums, 0, rotIdx-1, target);
}
return -1;
}
int binarySearch(vector<int> const &nums, int left, int right, int const target) {
while (left <= right) {
int mid = left + (right-left)/2;
if (nums[mid] == target) return mid;
if (target > nums[mid]) {
left = mid+1;
} else {
right = mid-1;
}
}
return -1;
}
int findRotIndex(vector<int> const &nums) {
int left = 0;
int right = nums.size()-1;
while (left <= right) {
int mid = left + (right-left)/2;
int rightVal = nums[mid+1];
if (rightVal < nums[mid]) {
return mid;
}
// 10 11 12 0 1 2 3
if (nums[mid] > nums[left]) {
left = mid+1;
} else {
right = mid;
}
}
return -1;
}
};
文档信息
- 本文作者：正义的伙伴
- 本文链接：https://ableasd.github.io/2021/04/17/LeetCode/
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