华为笔试 c++ AC 2.6 代码

字符串拆分 ac

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    vector<string> strs(n);
    for(int i = 0 ;i < n;i++)
    {
        cin >> strs[i];
    }
    multiset<string> ans;
    for(int i = 0;i < n;i++)
    {
        int nowSize = strs[i].size();
        //需要拆分
        if(nowSize > 8){
            int index = 0;
            while(nowSize > 8)
            {
                string now = strs[i].substr(index, 8);
                ans.insert(now);
                index += 8;
                nowSize -= 8;
            }
            string zero = "";
            for(int j = 0;j < 8 - nowSize;j++){
                zero += "0";
            }
            string now = strs[i].substr(index,nowSize);
            ans.insert(now + zero);
        }
        else{
            string zero = "";
            string now = strs[i].substr(0,nowSize);
            for(int j = 0;j < 8 - nowSize;j++){
                zero += "0";
            }
            ans.insert(now + zero);
        }

    }
    for(auto it = ans.begin(); it != ans.end();it++)
        cout << *it <<  " ";
    return 0;
}

字符串括号 ac

LeetCode 394 原题改编

#include<bits/stdc++.h>
using namespace std;
string decodeString(string s) {
        string res = "", t = "";
        stack<int> s_num;
        stack<string> s_str;
        int cnt = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] >= '0' && s[i] <= '9') {
                cnt = 10 * cnt + s[i] - '0';
            } else if (s[i] == '(' || s[i] == '[' || s[i] == '{') {
                s_num.push(cnt);
                s_str.push(t);
                cnt = 0; t.clear();
            } else if (s[i] == ')' || s[i] == ']' || s[i] == '}') {
                int k = s_num.top(); s_num.pop();
                for (int j = 0; j < k; ++j) s_str.top() += t;
                t = s_str.top(); s_str.pop();
            } else {
                t += s[i];
            }
        }
        return s_str.empty() ? t : s_str.top();
}

int main()
{
    string str;
    cin >> str;
    str = decodeString(str);
    for(int i = str.size() - 1;i >= 0;i--)
        cout << str[i];
    cout << endl;
}

海拔 63%样例

dfs遍历，注意边界条件。

#include<bits/stdc++.h>
using namespace std;
long long maxnum = 1e9;
void DFS(vector<vector<long long> > &grid, vector<vector<bool> > &visited, int x, int y, int z, int w, int& ans) {
    if (x < 0 || x >= grid.size()) return;
    if (y < 0 || y >= grid[0].size()) return;
    if(x == z && y == w){
        ans++;
        return;
    }
    int now = grid[x][y];
    if (visited[x][y]) return;

    visited[x][y] = true;
    if(x - 1 >=0 && grid[x - 1][y] > now )DFS(grid, visited, x - 1, y, z, w, ans);
    if(x + 1 < grid.size() && grid[x + 1][y] > now )DFS(grid, visited, x + 1, y, z, w, ans);
    if(y - 1 >= 0 && grid[x][y - 1] > now )DFS(grid, visited, x, y - 1, z, w, ans);
    if(y + 1 <= grid[0].size() && grid[x][y + 1] > now )DFS(grid, visited, x, y + 1, z, w, ans);
    visited[x][y] = false;
}
int main()
{
    int n,m;
    cin >> n >> m;
    vector< vector<long long> > g(n, vector<long long>(m, 0));
    vector<vector<bool> > visited(n, vector<bool>(m, false));

    for(int i = 0;i < n;i++){
        for(int j = 0; j < m;j++)
            cin >> g[i][j];
    }
    int x,y,z,w;
    cin >> x >> y >> z >> w;
    int ans = 0;
    DFS(g, visited, x, y, z, w, ans);
    ans = ans%maxnum;
    cout << ans << endl;
    return 0;
}

有没有第三题ac的大佬啊，dfs时间复杂度太高了。