int n, m, k; bool vis[N]; // vector<bool>vis(N); int ans = 0; vector<...
class Solution { public: vector<int> singleNumber(vector<int>& nums) { //使用图来实现哈希算法 map<int ,int> hash; //答案容器 vector<int> ans; //遍历两次,干脆解决 for(int i :nums){ hash[i]++; } for(auto [num,cnt]:hash){ if(cnt == 1) ans.push_back(num); } return ans; ...
int size_row = vec.size(); //获取行数 int size_col = vec[0].size(); //获取列数 4给vector二维数组赋值 简单的就直接赋值 ans[0][0]=1; ans[0][1]=2; ans[1][0]=3; ans[1][1]=4;
for ( int i=0; i<n; ++i ) a[i] = b[10+i] + c[20+i]; Use array notation where our operations on arrays do not require a specific order of operations among elements of the arrays. Specifying array sections An array section operator is written as one of the following: [first:le...
vector(const vector &ans);拷贝构造函数 代码示例：void printVector(vector<int>& v) { for (...
intmain{for(inti =1;i <=4;++i){intx;scanf("%d",&x);++cnt[x];}intans =0;for(inti =1;i <=4;++i) ans += cnt[i]/2;printf("%d\n",ans);return0;} B代码 // 最小的满足://x>= d//x%q = r// 若 d%q <= r// 则答案为 d + (r-d%q) ...
class Solution {public:int singleNumber(vector<int>& nums) {map<int,int> a;int n=nums.size();int ans=0;for(int i=0;i<n;i++){a[nums[i]]++;}for(int i=0;i<n;i++){if(a[nums[i]]==1){ans=nums[i];break;}}return ans;}}; ...
比如vector<int> ans; int t =accumulate(ans.begin(), ans.end(), 2),则t = sum{ans} + 2。比如vector ans = {"tt", "xx"}; string s =accumulate(ans.begin(), ans.end(), string(""));则s = "ttxx"。注意,如果val和vector中所装元素类型不同,会将vector里的元素转化成val的类型。
int i = 1;vector<int> nums = { 3,2,1 };int ans = 0;if (i - nums.size() < 0)ans = 1;return ans;我们想要ans返回的1,但时实际输出为0,通过调试我们发现，if语句⽆法进⼊，这就是因为i为int类型，⽽nums.size()为unsigned int类型，两者做运算，会强制更改为unsigned int类型，也就...
int ans; void dfs(int u) { for(int i=0;i<g[u].size();i++) { ans++; dfs(g[u][i]); } } int main() { int n; while(scanf("%d",&n)!=EOF) { memset(vis,0,sizeof(vis)); ans=0; int a; for(int i=0;i<=n;i++) ...