33. Search in Rotated Sorted Array
有一個按升序排序的整數數組 nums（具有不同的值）。
在傳遞給你的函數之前，nums 可能在未知的旋轉點 k 處進行了旋轉（1 <= k < nums.length），使得結果數組為 [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]（0 索引）。例如，[0,1,2,4,5,6,7] 可能在旋轉點索引 3 處進行旋轉，變為 [4,5,6,7,0,1,2]。
給定可能旋轉後的數組 nums 和一個整數 target，如果 target 在 nums 中，則返回 target 的索引；如果不在 nums 中，則返回 -1。
你必須設計一個時間複雜度為 O(log n) 的算法。
範例 1：
輸入：nums = [4,5,6,7,0,1,2], target = 0 輸出：4
Python
from typing import List
class Solution:
def search(self, nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
# Determine which part is sorted
if nums[left] <= nums[mid]:
# Left part is sorted
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
# Right part is sorted
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
return -1
16.81MB, 48ms
C++
#include <vector>
using namespace std;
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
}
// Determine which part is sorted
if (nums[left] <= nums[mid]) {
// Left part is sorted
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
// Right part is sorted
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
}
};
13.35MB, 4ms
Javascript
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function(nums, target) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (nums[mid] === target) {
return mid;
}
// Determine which part is sorted
if (nums[left] <= nums[mid]) {
// Left part is sorted
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
// Right part is sorted
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
};
48.28MB, 68ms
