就只想到了O(n)的，
想到的二分，
- target<nums[left] 则在右边有序搜索
- target>=nums[left],在左边有序搜素
- 自然是错误的，于是搁置不得了之。
看了题解恍然大悟。
以mid分成左右两个数组。
其中至少一个是有序的，且范围已知
- 如果target在此范围内，二分即可
- 不在此范围内，搜索另一个数组
class Solution {
public int search(int[] nums, int target) {
return searchHelp(nums,target,0,nums.length-1);
}
private int searchHelp(int[] nums,int target,int left,int right){
if(left>right) return -1;
int mid=left+((right-left)>>1);
if(nums[mid]==target) return mid;
if(mid-1>=left&&nums[left]<=nums[mid-1]){
//有序在左边，且目标值在里面
if(target>=nums[left]&&target<=nums[mid-1]){
right=mid-1;
while (left<=right){
mid=left+(right-left)/2;
if(nums[mid]==target) return mid;
if(nums[mid]>target) right=mid-1;
else left=mid+1;
}
}//不在里面，搜索右边无序的。
return searchHelp(nums,target,mid+1,right);
}else {
//有序在右边，且目标在里面
if(mid+1<nums.length&&target<=nums[right]&&target>=nums[mid+1]){
left=mid+1;
while (left<=right){
mid=left+(right-left)/2;
if(nums[mid]==target) return mid;
if(nums[mid]>target) right=mid-1;
else left=mid+1;
}
}
return searchHelp(nums,target,left,mid-1);
}
}
}
还有一种更简洁的写法
class Solution {
public int search(int[] nums, int target) {
int left=0;
int right=nums.length-1;
while (left<=right){
int mid=left+((right-left)>>1);
if(nums[mid]==target) return mid;
if(mid-1>=left&&nums[left]<=nums[mid-1]){
if(target<=nums[mid-1]&&target>=nums[left]){
right=mid-1;
}else left=mid+1;
}else{
if(mid+1<nums.length&&target>=nums[mid+1]&&target<=nums[right]){
left=mid+1;
} else right=mid-1;
}
}
return -1;
}
}
关于判断
if(mid-1>=left&&nums[left]<=nums[mid-1]){
nums[left]<=nums[mid-1] 取等号，是为了考虑只有一个数的时候