題目敘述
There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Constraints:
1 <= nums.length <= 5000-10**4 <= nums[i] <= 10**4nums is guaranteed to be rotated at some pivot.-10**4 <= target <= 10**4
Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
題目翻譯
今天有一個排序過的陣列 nums，但會被挪移。在這挪移過後的陣列中找是不是含有 target 存在。如果存在就回傳 true，沒有就回傳 false。
解法解析
這題是使用 Binary Search，但要注意的是其開頭和結尾是切開來的。細節的話可以看看官網本身的講解，我覺得滿清楚的 - Solution
解法範例
Go
func search(nums []int, target int) bool {
n := len(nums)
if n == 0 {
return false
}
end, start := n-1, 0
for start <= end {
mid := (start + end) / 2
if nums[mid] == target {
return true
}
if nums[start] == nums[mid] {
start += 1
continue
}
if nums[mid] > nums[start] {
if nums[mid] > target && target >= nums[start] {
end = mid - 1
} else {
start = mid + 1
}
} else {
if target > nums[mid] && nums[end] >= target {
start = mid + 1
} else {
end = mid - 1
}
}
}
return false
}
JavaScript
var search = function (nums, target) {
const n = nums.length;
if (n === 0) return false;
let end = n - 1,
start = 0;
while (start <= end) {
let mid = Math.floor((start + end) / 2);
if (nums[mid] === target) return true;
if (nums[start] === nums[mid]) {
start++;
continue;
}
if (nums[mid] > nums[start]) {
if (nums[mid] > target && target >= nums[start]) {
end = mid - 1;
} else {
start = mid + 1;
}
} else {
if (target > nums[mid] && nums[end] >= target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
}
return false;
};
Kotlin
PHP
class Solution
{
function search($nums, $target)
{
$n = count($nums);
if ($n == 0) {
return false;
}
$end = $n - 1;
$start = 0;
while ($start <= $end) {
$mid = intdiv($start + $end, 2);
if ($nums[$mid] == $target) {
return true;
}
if ($nums[$start] == $nums[$mid]) {
$start++;
continue;
}
if ($nums[$mid] > $nums[$start]) {
if ($nums[$mid] > $target && $target >= $nums[$start]) {
$end = $mid - 1;
} else {
$start = $mid + 1;
}
} else {
if ($target > $nums[$mid] && $nums[$end] >= $target) {
$start = $mid + 1;
} else {
$end = $mid - 1;
}
}
}
return false;
}
}
Python
class Solution:
def search(self, nums: List[int], target: int) -> bool:
n = len(nums)
if n == 0:
return False
end = n - 1
start = 0
while start <= end:
mid = (start + end) // 2
if nums[mid] == target:
return True
if nums[start] == nums[mid]:
start += 1
continue
pivot_array = nums[start] <= nums[mid]
if pivot_array ^ (nums[start] <= target):
if pivot_array:
start = mid + 1
else:
end = mid - 1
else:
if nums[mid] < target:
start = mid + 1
else:
end = mid - 1
return False
Rust
Swift