LeetCode 之 Search for a Range — C++ 实现

Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return 

[-1, -1]

.

For example,

Given 

[5, 7, 7, 8, 8, 10]

 and target value 8,

return 

[3, 4]

.
给定一个有序的整型数组，找出给定目标值在数组中的起始位置和终止位置。
算法时间复杂度要求为 O(logn).

如果目标值未在数组中，返回[-1,
-1]  

例如，

给定数组[5,
7, 7, 8, 8, 10] 和目标值 8，返回[3,
4]。

分析：

二分查找。

实现1：

    先用二分法找到一个值，然后在这个值的左、右利用上次的上(high)下(low)界，分别在 low 和 mid 以及 mid 和 high 之间再进行二分查找。

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int low = 0, high = nums.size()-1;
int mid = 0, lmid = 0, rmid = 0;
int  left = 0, right = 0;
int isfind = 0; //存在标志

vector<int> returnSize(2, -1);
if(nums.empty()) //空指针或空数组
{
return returnSize;
}

while(low <= high) //二分查找
{
mid = (low + high)/2;
if(nums[mid] == target) //找到一个匹配节点
{
isfind = 1;
break;
}
else if(nums[mid] < target)
{
low = mid + 1;
}
else
{
high = mid - 1;
}
}

if(isfind)
{
left = mid;
while(low <= left)//二分找左边界
{
lmid = (low + left)/2;
if(nums[lmid] == target)
{
if(lmid-1 >= low)
{
if(nums[lmid-1] == target)
{
left = lmid - 1;
}
else
{
returnSize[0] = lmid;
break;
}
}
else
{
returnSize[0] = lmid;
break;
}
}
else
{
low = lmid + 1;
}
}

right = mid;
while(right <= high)//二分找右边界
{
rmid = (right + high)/2;
if(nums[rmid] == target)
{
if(rmid+1 <= high)
{
if(nums[rmid+1] == target)
{
right = rmid + 1;
}
else
{
returnSize[1] = rmid;
break;
}
}
else
{
returnSize[1] = rmid;
break;
}
}
else
{
high = rmid - 1;
}
}
}

return returnSize;

}
};

实现2：

     对左、右范围位置分别进行二分查找。查找左范围时，当mid位置的值小于等于 target 时，向左边查找；查找又范围时，当mid值大于等于 target 时，向右边查找。

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int low = 0, high = nums.size()-1;
int lmid = 0, rmid = 0;

vector<int> returnSize(2, -1);
if(nums.empty()) //空指针或空数组
{
return returnSize;
}

while(low <= high) //对左边界进行二分查找
{
lmid = (low + high)/2;
if(nums[lmid] < target)
{
low = lmid + 1;
}
else
{
high = lmid - 1;
}
}
if(nums[low] != target)//没找到
{
return returnSize;
}
else
{
returnSize[0] = low;
}

low = 0;
high = nums.size()-1;
while(low <= high) //对右边界进行二分查找
{
rmid = (low + high)/2;
if(nums[rmid] <= target)
{
low = rmid + 1;
}
else
{

4000
high = rmid - 1;
}
}
returnSize[1] = high;

return returnSize;
}
};