标签：ted 最大值 特殊 nbsp while fir ges size def
11. 盛最多水的容器
class Solution { public: int maxArea(vector<int>& height) { vector<int> & v =height; int l =0,r=v.size()-1; int res = 0; while(l<r) { int lval = v[l]; int rval = v[r]; int minval = min(lval,rval); res =max(res,minval*(r-l)); if(lval<rval) l++; else r--; } return res ; } };
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> v(2,-1); int high =binarySearchUp(nums,target,0,nums.size()-1); int low =binarySearchDown(nums,target,0,nums.size()-1); //DBG2(high,low); if(high >= low ) { v[0]=low; v[1]=high; } return v; } private: int binarySearchUp(vector<int> & nums ,int target,int begin,int end) { if(begin>end)return end; int mid =(begin+end) >>1; if(nums[mid]>target) return binarySearchUp(nums,target,begin,mid-1); else return binarySearchUp(nums,target,mid+1,end); } int binarySearchDown(vector<int> & nums ,int target,int begin,int end) { if(begin>end)return begin; int mid =(begin+end) >>1; if(nums[mid]<target) return binarySearchDown(nums,target,mid+1,end); else return binarySearchDown(nums,target,begin,mid-1); } };
59. 螺旋矩阵 II
class Solution { public: vector<vector<int> > generateMatrix(int n) { vector<vector<int> > v (n,vector<int> (n,0)); int col_min=0,col_max=n-1,row_min=0,row_max=n-1; int row=-1,col=0; int i=1; while(i<=n*n) { while(++row<=row_max){ v[col][row]=i++; } row--; while(++col<=col_max) { v[col][row]=i++; } col --; row_max--; while(--row>=row_min) { v[col][row]=i++; } row++; col_max--; col_min++; while(--col>=col_min) { v[col][row]=i++; } col++;row_min++; } return v; } };
60. 第k个排列
class Solution { public: string getPermutation(int n, int k) { vector<int> v; for(int i=1;i<=n;i++) v.push_back(i); for(int i=0;i<k;i++) { next_permutation(v.begin(),v.end()); }; string str =""; for(auto i :v) { str+=i; } return str; } };
141. 环形链表
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool hasCycle(ListNode *head) { if(head==NULL) return false; ListNode * fast=head,*last=head; if(fast->next&& fast->next->next) fast =fast->next->next; else return false ; if(last->next)last=last->next; while(fast!=last) { if(fast->next&& fast->next->next) fast =fast->next->next; else return false ; if(last->next)last=last->next; } return true; } };
class MinStack { public: /** initialize your data structure here. */ stack<int>all; stack<int>minstack; MinStack() { } void push(int x) { if(all.empty()) { all.push(x); minstack.push(x); } else { all.push(x); if(minstack.top()>=x)minstack.push(x);; } } void pop() { if(all.top()==minstack.top()) minstack.pop(); all.pop(); } int top() { return all.top(); } int getMin() { return minstack.top(); } }; /** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */
416. Partition Equal Subset Sum
class Solution { public: bool canPartition(vector<int>& nums) { int sum = accumulate(nums.begin(), nums.end(), 0), target = sum >> 1; if (sum & 1) return false; vector<bool> dp(target + 1, false); dp[0] = true; for (int num : nums) { for (int i = target ; i >=num ; --i) { dp[i] = dp[i] || dp[i - num]; } } return dp[target]; } };
// Forward declaration of guess API. // @param num, your guess // @return -1 if my number is lower, 1 if my number is higher, otherwise return 0 int guess(int num); class Solution { public: int guessNumber(int n) { int l=1,r=n; int mid,ret; while(l<r) { mid = l/2+r/2; // mid=l+r>>1; ret = guess(mid); if(ret==0) return mid; else if(ret< 0 ) r=mid-1; else l=mid+1; } return l; } };
378. 有序矩阵中第K小的元素
class Solution { public: int kthSmallest(vector<vector<int>>& matrix, int k) { priority_queue<int> q ; for(auto & v : matrix) for(auto i:v) { q.push(i); if(q.size()>k)q.pop(); } return q.top(); } };
class Solution { public: bool canConstruct(string ransomNote, string magazine) { // unordered_map<char,int> a; int a[256] ={0}; for(auto i:magazine) { a[i]++; } for(auto i:ransomNote) { if(a[i]> 0 ) a[i]--; else return false; } return true; } };
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515. 在每个树行中找最大值
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> largestValues(TreeNode* root) { vector<int> v; if(root==NULL) return v; queue<TreeNode*> q ; q.push(root); while(1) { int tmp = getnum(q); v.push_back(tmp); getnxt(q) ; if(q.empty())break; } return v; } void getnxt(queue<TreeNode*>& tmp ) { queue<TreeNode*> t ; while(!tmp.empty()) { TreeNode* i=tmp.front(); if(i->left) { t.push(i->left); } if(i->right) { t.push(i->right); } tmp.pop(); } while(!t.empty()) tmp.push(t.front()),t.pop(); } int getnum(queue<TreeNode*> tmp) { int mm = -2147483648; while(!tmp.empty()) { int tt = tmp.front()->val; mm=max(mm,tt); tmp.pop(); } return mm; } };
class Solution { public: int findLUSlength(string a, string b) { return a==b?-1:max(a.size(),b.size()); } };
540. 有序数组中的单一元素
class Solution { public: int singleNonDuplicate(vector<int>& nums) { int ans =0; for(auto i:nums) ans^=i; return ans; } };
639. 解码方法 2
class Solution { public: int ways(char c) { if (c == ‘*‘) return 9; if (c == ‘0‘) return 0; return 1; } int ways(char i, char j) { if (i == ‘*‘ && j == ‘*‘) return 15; if (i == ‘*‘) return (j <= ‘6‘ ? 2 : 1); if (j == ‘*‘) { if (i == ‘1‘) return 9; if (i == ‘2‘) return 6; return 0; } if (i == ‘1‘ || (i == ‘2‘ && j <= ‘6‘)) return 1; return 0; } int numDecodings(string s) { int n =s.length(); if(n==0||s[0]==‘0‘) return 0; long f2=1,f1= (long)ways(s[0]); long res = f1; long mod = (long)1e9+7; for(int i=1;i<n;i++) { res = ways(s[i-1],s[i])*f2 + ways(s[i])*f1; res%=mod; f2=f1; f1= res; } return res; } };
class MapSum { public: /** Initialize your data structure here. */ map<string,int> m; MapSum() { } void insert(string key, int val) { m[key] =val; } int sum(string prefix) { int s = 0; for (auto one : m) { if (prefix == one.first.substr(0, prefix.length())) s += one.second; } return s; } }; /** * Your MapSum object will be instantiated and called as such: * MapSum obj = new MapSum(); * obj.insert(key,val); * int param_2 = obj.sum(prefix); */
743. 网络延迟时间
#define REP(i,l,r) for(int i=l;i<=r;i++) class Solution { public: int networkDelayTime(vector<vector<int>>& times, int N, int K) { const int inf = (int)1e5+9; int n=N,k=K; int m[n+1][n+1]; int dist[n+1],vis[n+1]; REP(i,1,n) REP(j,1,n) if(i==j)m[i][j] ;else m[i][j]=inf; for(auto i:times)m[i[0]][i[1]]=i[2]; REP(i,1,n) dist[i]=m[k][i],vis[i]=0; vis[k]=1; int ans =-1; REP(i,1,n-1) { int minx=inf,u=k; REP(j,1,n) if(!vis[j] && dist[j]!=inf &&dist[j]<minx) minx=dist[j],u=j; if(minx==inf) return -1; vis[u]=1; ans=max(ans,minx); REP(j,1,n) if(!vis[j]&& dist[j]>dist[u]+m[u][j]) dist[j]=dist[u]+m[u][j]; } return ans; } };
标签：ted 最大值 特殊 nbsp while fir ges size def
原文地址：https://www.cnblogs.com/corx/p/9001830.html
