LeetCode（34）Search for a Range

题目

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].

分析

给定一有序整数序列，与目标元素值，要求输出目标元素值在此序列中出现的范围下标。且复杂度控制在O(logn)内。

明显的，我们应该采取二分搜索的思想，设计求出关键字最早出现位置与最后出现位置，与普通的二叉搜索比较，只需要修改判断条件即可。

AC代码

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ret;
if (nums.size() == 0)
{
ret.push_back(-1);
ret.push_back(-1);
return ret;
}

//寻找目标元素的下标
int pos = BinarySearch(nums, target);

//目标元素不存在
if (pos == -1)
{
ret.push_back(-1);
ret.push_back(-1);
return ret;
}
else{
int left = BinarySearchLeft(nums, 0, pos, target);
int right = BinarySearchRight(nums, pos, nums.size()-1 , target);
ret.push_back(left);
ret.push_back(right);
return ret;
}//if

}

int BinarySearch(vector<int> & nums, int target)
{
int left = 0, right = nums.size() - 1;

while (left <= right)
{
int mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
else if (nums[mid] < target)
{
left = mid + 1;
continue;
}
else{
right = mid - 1;
continue;
}
}//while

return -1;
}

int BinarySearchLeft(vector<int> & nums, int left, int right, int target)
{
while (left < right)
{
int mid = (left + right) / 2;
if (nums[mid] == target && nums[mid-1] < target)
return mid;
else if (nums[mid] < target)
{
left = mid + 1;
continue;
}
else{
right = mid - 1;
continue;
}
}//while
return left;
}

int BinarySearchRight(vector<int> & nums, int left, int right, int target)
{
while (left < right)
{
int mid = (left + right) / 2;
if (nums[mid] == target && nums[mid + 1] > target)
return mid;
else if (nums[mid] <= target)
{
left = mid + 1;
continue;
}
else{
right = mid - 1;
continue;
}
}//while
return right;
}
};

GitHub测试程序源码