LeetCode 33.Search in Rotated Sorted Array（M）
2024-09-07 08:09:19
题目：
Suppose an array sorted in ascending order 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).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
思路：
1.当数组长度为0时，return -1；
2.当数组长度为1时，比较nums[0]与target关系;
3.当数组长度大于1，且为rotated sorted array时，根据nums[mid]与nums[start]关系判断如何移动；
4.当数组长度大于1，但不是rotated sorted array时，用传统二分法；
5.注意，不仅仅要比较target与nums[mid]关系，还要比较其与nums[start]和nums[end]关系，否则会出现死循环。
代码：
public class Solution {
public int search(int[] nums, int target) {
int start = 0,end = nums.length-1,mid = 0;
if(nums.length == 0){
return -1;
}
if(nums.length == 1){
if(nums[0] == target){
return 0;
}else{
return -1;
}
}
while(nums[start] > nums[end] && start + 1 < end){
mid = start + (end - start)/2;
if(nums[mid] == target){
return mid;
}else if(nums[mid] > nums[start]){
if(nums[mid] <= target ){
start = mid;
}else if(nums[start] <= target){
end = mid;
}else if(nums[start] >= target){
start = mid;
}
}else if(nums[mid] < nums[start]){
if(nums[mid] >= target){
end = mid;
}else if(nums[start] <= target){
end = mid;
}else if(nums[start] >= target){
start = mid;
}
}
}
while(nums[start] < nums[end] && start + 1 < end){
mid = start + (end - start)/2;
if(nums[mid] == target){
return mid;
}else if(nums[mid] >= target){
end = mid;
}else{
start = mid;
}
}
if(nums[start] == target){
return start;
}else if(nums[end] == target){
return end;
}
return -1;
}
}
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