刷题心得 leetcode 704 leetcode 27

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风格）

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);
    // 非常量引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
这边文章解释的很清楚了，非常量引用传参进去后可能会被修改，如果是个右值显然是不行的

再次运行报不可以初始化，原来vscode默认只支持c++98标准，在.vscode里的task.json添加”-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。
可以看出为什么api之类的都采用左闭右开，写出来的代码更自然，计算区间什么的也更直接。

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();
    }
};

快慢指针版本:
算法思路参考文章:
https://programmercarl.com/0027.%E7%A7%BB%E9%99%A4%E5%85%83%E...

class Solution
{
public:
    int removeElement(vector<int> &nums, int val)
    {
        int slow = 0;
        for (int fast = 0; fast < nums.size(); ++fast)
        {
            if (nums[slow] != val)
            {
                nums[slow] = nums[fast];
                ++slow;
            }
        }

        return slow;
    }
};