以前总觉得二分查找简单,刷了一些题目后再也不觉得二分查找简单了,各种边界条件,简直要了老命,但对这道题而言,我们只要找到某半段有序数组即可,如果target恰好位于其中,那么问题就转换为普通的二分查找。下面是两种形式的写法:
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0,r = nums.size() - 1;
while(l <= r)
{
int mid = l + (r - l)/2;
if(nums[mid] == target) return mid;
if(nums[mid] < nums[r])
{
//后半段有序
if(nums[mid] < target && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}else{
//前半段有序
if(nums[l] <= target && target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}
}
return -1;
}
};
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0,r = nums.size() - 1;
while(l <= r)
{
int mid = l + (r - l)/2;
if(nums[mid] == target) return mid;
if(nums[mid] > nums[r])
{
//前半段有序,且target位于前半段
if(nums[l] <= target && target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}else{
//后半段有序,且target位于后半段
if(nums[mid] < target && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}
}
return -1;
}
};
2021.06.10更新另一种思路:
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0,r = nums.size() -1;
while(l <= r)
{
int mid = l + (r-l)/2;
if(nums[mid] == target) return mid;
//[3,4,5,1,2]
if(nums[mid] > nums[r])
{
if(target < nums[mid] && target >= nums[l])
r = mid - 1;
else
l = mid + 1;
//[5,1,2,3,4]
}else if(nums[mid] < nums[l]){
if(target > nums[mid] && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}else{
if(target < nums[mid])
r = mid -1;
else
l = mid + 1;
}
}
return -1;
}
};
仔细分析上面两种代码解法,实际是上是一样的。