0 简单 c++转 js 代码

1. 函数头部返回值=>function, 参数列表去类型
2. int|double|string >= var
3. .size() => .length
4. / 2 => >>1

function cpp2js(str){
    var from = str.split('\n'), to = [];
    for(var s of from){
        var out = s;
        var func = /\w+\s+(\w+)\s*\(((<.*>|[^<>])*)\)/
        if(func.test(s)){
            out = "function ";
            out += RegExp.$1 + "("
            out += RegExp.$2.replace(/([^, ]+\s+)+(\w+)/g, "$2")
            out += ")"
            out += RegExp.$3 || ''
        }
        if(/\.size/.test(out)){
            out = out.replace(/\.size\(\)/,'.length');
        }
        if(/\/\s*2/.test(out)){
            out = out.replace(/\/\s*2/,">>1");
        }
        if(/int|double|string/.test(out)){
            out = out.replace(/int|double|string/,"var")
        }
        to.push(out);
    }
    return to.join("\n")
}

1 函数体内运算

1.1 lamda 表达式

[1,2,3].map(x=>x+1) //js
[x+1 for x in [1,2,3]] //python

1.2 解构赋值-交换

[ a , b ] = [ b, a+b ] //js
a , b  = b, a+b  //python
std::swap(a,b) //c++

1.3 运算符

&& || ! 'i>=0?a[i]:0'//js,c++
and or not 'a[i] if i>=0 else 0' //python

2 数组/List/vector操作

2.1 获取长度

a.length //js
len(a) //python
a.size() //c++

2.2 尾部 push 和 pop

a.push(1), b = a.pop() //js
a.append(1) , b = a.pop() //python
a.push_back(1),  b=a.back();a.pop_back()//c++

2.3 排序

//升序 , 降序
a.sort((a,b)=>a-b) , a.sort((a,b)=>b-a)// js
a.sort(), a.sort(reverse=True) //python
std::sort(a.begin(),a.end()); std::sort(a.rbegin(),a.rend()) //c++

2.4 遍历

1. 拿到下标                    2.拿到值
//js
for(let i in a)               for(let i of a) 
//python
for i in range(len(a)):       for i in a:
//c++
int b[] = {2,6,5,4,3,1};
vector<int> a(b,b+6);
for(auto i = a.begin();i!=a.end();i++){
   cout<< *i;
}
for(auto i : a){
  cout<<i;
}

3 字符串操作

4 算法举例

4.1 简单快排=>快速搜索第 K 大数字

# js
function f(a,k){ return a.sort((a,b)=>b-a)[k-1] } 
# python
class Solution:
    def f(self, nums: 'List[int]', k: 'int') -> 'int':
        nums.sort(reverse=True)
        return nums[k-1]
# c++
class Solution {
public:
    int f(vector<int>& nums, int k) {
        std::sort(nums.rbegin(),nums.rend());
        return nums[k-1];
    }
};

4.2 手撸快排=>快速搜索第 K 大数字

# js
# 和 下面 C++的基本一致
# python
class Solution:
class Solution:
    def findKthLargest(self, nums: 'List[int]', k: 'int') -> 'int':
        l,r = 0,len(nums)-1
        while l<=r:
            p = self.partition(nums,l,r)
            if p==k-1: return nums[p]
            elif p>k-1: r = p-1
            else: l = p+1
        return -1
    def partition(self,nums,l,r):
        k = nums[l]
        while l<r:
            while l<r and nums[r]<=k:r=r-1
            nums[l] = nums[r]
            while l<r and nums[l]>=k:l=l+1
            nums[r] = nums[l]
        nums[r] = k
        return l
# c++
class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        int l=0,r=nums.size()-1;
        while(l<=r){
              int p = partition(nums,l,r);
              if(p == k-1) return nums[p];
              if(p<k-1) l = p+1;
              else r = p-1;
        }
        return -1;
    }
    int partition(vector<int>&nums,int l, int r){
        int k = nums[l];
        while(l<r){
           while(l<r && nums[r] <= k) r--;
           nums[l] = nums[r];
           while(l<r && nums[l] >= k) l++;
           nums[r] = nums[l];
        }
        nums[l] = k;
        return l;
    }
};