【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
Search for a Range
Given an array of integers nums sorted in ascending order, 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].
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
/*问题：在有序数组中查找目标值出现的范围
方法：对于有序序列用二分查找
使用stl中lower_bound和upper_bound函数
O(logn)
*/
class Solution
{
public:
vector<int> searchRange(vector<int>& nums, int target)
{
if(nums.empty()) return vector<int>{-1,-1}; //容器为空时 （vector<int>{}为初始化返回容器操作）
//lower_bound找到后返回第一个大于等于target的位置的迭代器，否则返回迭代器end
vector<int>::iterator start = lower_bound(nums.begin(), nums.end(),target);
//upper_bound找到后返回第一个大于target的位置的迭代器，否则返回迭代器end
vector<int>::iterator end = upper_bound(nums.begin(), nums.end(), target);
if(start == end || start == nums.end())//相等时，说明找到的是大于目标值的数，指向end时说明目标值比数组所有元素大，两种情况下均为没有找到
return {-1,-1};
else
return {start-nums.begin(), end-nums.begin()-1};
}
};
/*
//或者自己实现
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> result;
int begin = 0;
int end = nums.size() - 1;
int start_index = -1;
int end_index = -1;
//locate the start_index
while (begin <= end){
int mid = (begin + end) / 2;
if (nums[mid] == target){
if (mid - 1<0 || nums[mid - 1] != target){
start_index = mid;
break;
}
else{
end = mid - 1;
}
}
else if (nums[mid]>target){
end = mid - 1;
}
else{
begin = mid + 1;
}
}
//locate the end_index
begin = 0;
end = nums.size() - 1;
while (begin <= end){
int mid = (begin + end) / 2;
if (nums[mid] == target){
if (mid + 1>nums.size() - 1 || nums[mid + 1] != target){
end_index = mid;
break;
}
else{
begin = mid + 1;
}
}
else if (nums[mid]>target){
end = mid - 1;
}
else{
begin = mid + 1;
}
}
//cout << start_index << " " << end_index << endl;
result.push_back(start_index);
result.push_back(end_index);
return result;
}
};
*/