- Search in Rotated Sorted Array
Medium
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You are given an integer array nums
sorted in ascending order, and an integer target
.
Suppose that nums
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]
).
If target
is found in the array return its index, otherwise, return -1
.
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
nums
are unique. nums
is guranteed to be rotated at some pivot.-10^4 <= target <= 10^4
主要代码
class Solution {
public int search(int[] nums, int target) {
if (nums.length == 1) {
if (nums[0] == target) {
return 0;
} else {
return -1;
}
}
//1. 找到分割点
int len = nums.length;
int tmp = 0;
for (int i = 1; i < len; i++) {
if (nums[i] < nums[i - 1]) {
tmp = i;
}
}
int l = 0, r = tmp - 1;
while (l < r) {
int mid = (l + r) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > target) {
r = mid - 1;
} else {
l = mid + 1;
}
}
if (nums[l] == target) {
return l;
}
l = tmp;
r = len - 1;
while (l < r) {
int mid = (l + r) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > target) {
r = mid - 1;
} else {
l = mid + 1;
}
}
if (nums[l] == target) {
return l;
}
return -1;
}
}