leetcode 704 题目链接：
https://leetcode.com/problems/binary-search/
尝试 c++ 写题，搭建架子代码（尝试搭建ACM格调）
class Solution{public: int search(vector<int> &nums, int target) { }};int main(){ Solution sol; vector<int> nums = {-1, 0, 3, 5, 9, 12}; sol.search({-1, 0, 3, 5, 9, 12}, 9); // 十分量援用的初始值必须为左值 // initial value of reference to non-const must be an lvalue}https://blog.csdn.net/hou09tian/article/details/80565343
这边文章解释的很分明了，十分量援用传参进去后可能会被批改，如果是个右值显然是不行的
再次运行报不能够初始化，原来vscode默认只反对c++98规范，在.vscode里的task.json增加”-std=c++11“编译参数即可
算法思路文章参考：
https://programmercarl.com/0704.%E4%BA%8C%E5%88%86%E6%9F%A5%E...
最终ac代码：（左闭右开）
#include <iostream>#include <vector>using namespace std;class Solution {public: int search(vector<int>& nums, int target) { size_t start = 0; size_t end = nums.size(); while (start < end) { size_t middle = start + ((end - start) >> 1); if (nums[middle] == target) { return static_cast<int>(middle); } else if (nums[middle] > target) { end = middle; } else { start = middle + 1; } } return -1; }};再次尝试左闭右闭：
class Solution{public: int search(vector<int> &nums, int target) { size_t start = 0; size_t end = nums.size() - 1; while (start <= end) { size_t middle = start + ((end - start + 1) >> 1); if (nums[middle] == target) { return static_cast<int>(middle); } else if (nums[middle] > target) { end = middle - 1; } else { start = middle + 1; } } return -1; }};[5], -5 的用例过不去，tle了，将size_t改为int，通过。
能够看出为什么api之类的都采纳左闭右开，写进去的代码更天然，计算区间什么的也更间接。
leetcode 27 题目链接:
https://leetcode.com/problems/remove-element/
ac代码:
class Solution{public: int removeElement(vector<int> &nums, int val) { for (vector<int>::iterator it = nums.begin(); it != nums.end();) { if (*it == val) { it = nums.erase(it);// O(n) time complexity } else { it++;// prefer pre-increment for post-increment making a temporary copy } } return nums.size(); }};