81. Search in Rotated Sorted Array II
給定一個按非遞減順序排序的整數數組 nums（不一定是唯一值）。
在傳遞給你的函數之前，nums 在未知的旋轉點 k（0 <= k < nums.length）處旋轉，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，[0,1,2,4,4,4,5,6,6,7] 可能在旋轉點索引 5 處旋轉並變為 [4,5,6,6,7,0,1,2,4,4]。
給定旋轉後的數組 nums 和一個整數 target，如果 target 存在於 nums 中則返回 true，否則返回 false。
你必須儘可能減少整體操作步驟。
範例 1：
輸入：nums = [2,5,6,0,0,1,2], target = 0 輸出：true
Python
from typing import List
class Solution:
def search(self, nums: List[int], target: int) -> bool:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return True
# If the left half is sorted
if nums[left] < nums[mid]:
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
# If the right half is sorted
elif nums[left] > nums[mid]:
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
# If nums[left] == nums[mid], we have to deal with duplicates
else:
left += 1
return False
17.32MB, 52ms
C++
#include <vector>
class Solution {
public:
bool search(std::vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return true;
}
// If the left half is sorted
if (nums[left] < nums[mid]) {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
}
// If the right half is sorted
else if (nums[left] > nums[mid]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
// If nums[left] == nums[mid], we have to deal with duplicates
else {
left++;
}
}
return false;
}
};
16.52MB, 3ms
Javascript
/**
* @param {number[]} nums
* @param {number} target
* @return {boolean}
*/
var search = function(nums, target) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] === target) {
return true;
}
// If the left half is sorted
if (nums[left] < nums[mid]) {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
}
// If the right half is sorted
else if (nums[left] > nums[mid]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
// If nums[left] == nums[mid], we have to deal with duplicates
else {
left++;
}
}
return false;
};
49.04MB, 48ms
