题解:
先判断nums[mid]是在旋转数组的左半边,还是右半边;
如果在左半边然后使用target和nums[0]和nums[mid]作比较,target处于[0,mid]中间,right = mid - 1; else left = mid + 1;
如果在右半边,使用target和nums[mid] nums[nums.length-1]作比较,target处于[mid,nums[nums.length-1]], left = mid + 1,否则right = mid-1
代码
public int search(int[] nums, int target) {
if(nums.length == 0){
return -1;
}
int left = 0, right = nums.length - 1;
while(left <= right){
int mid = left + (right - left) / 2;
if(nums[mid] == target){
return mid;
}
//左半边有序,在左半边使用二分查找
if(nums[mid] >= nums[0]){
if(nums[0] <= target && target < nums[mid]){ //target处于[0,mid),向左移动mid right = mid - 1;
}
else{
left = mid + 1;
}
}
//右半边有序,在右半边使用二分查找
else{
if(nums[mid] < target && target <= nums[nums.length - 1]){
left = mid + 1;
}
else{
right = mid - 1;
}
}
}
return -1;
}