description:
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].
给定一个值，找到在有序数组中的最左边和最右边出现的值的坐标
Example:
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]
answer:
https://blog.csdn.net/MC_007/article/details/80998686
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
if (nums.empty()) return {-1, -1};
if (target < nums[0] || target > nums[nums.size() - 1]) return {-1, -1};
const int l = distance(nums.begin(), lower_bound(nums.begin(), nums.end(), target));
const int u = distance(nums.begin(), prev(upper_bound(nums.begin(), nums.end(), target)));
// 讲真我觉得下面两句不应该这么写，因为如果list中所有值都小于target的话，那就会返回last，再用nums[last]就是不合法的，因为last是超出范围的
if (nums[l] != target) return {-1, -1}; //如果list中所有值都小于target，就返回last
else return {l, u};
}
};
my answer:
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
if (nums.empty()) return {-1, -1};
if (target < nums[0] || target > nums[nums.size() - 1]) return {-1, -1};
const int l = distance(nums.begin(), lower_bound(nums.begin(), nums.end(), target));
const int u = distance(nums.begin(), prev(upper_bound(nums.begin(), nums.end(), target)));
if (l == nums.size()) return {-1, -1}; // target bigger than all value in list
else{
if (nums[l] != target) return {-1, -1}; // target lower than all value in list
else return {l, u};
}
}
};
relative point get√：
- prev：返回指针的前一个值处的指针
- lower_bound：二分查找的函数有 3 个： 参考：C++ lower_bound 和upper_bound 【https://www.cnblogs.com/Tang-tangt/p/9291018.html】
- distance：俩个指针之间的距离