第十九周：[sicily] 算法机考模拟题

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1000. 函数求值

保存历层结果暴力破解

     int F(int k, int n) {           int a[k+2][n+2];           for(int i = 0; i < n+1; i++)                a[0][i] = i;           for (int i = 0; i < k; i++) {               int total = 0;               for (int j = 1; j <= n; j++) {                   total += a[i][j];                   a[i+1][j] = total;               }           }           return a[k][n];      }

1001.会议安排

对A、B增序排序，按顺序比较算出最大会议数

    int assignConferenceRoom(vector<int>& A, vector<int>& B) {       sort(A.begin(), A.end());       sort(B.begin(), B.end());       int res = 0;           for(int j = 0; res<A.size() && j < B.size();j++) {           if (A[res] <= B[j]) {               res++;           }       }       return res;          }

1002.等价二叉树

递归求解即可

   bool isEqual(TreeNode* p, TreeNode* q) {      if(p == NULL && q == NULL)          return true;      if(p == NULL || q == NULL)          return false;      return (p->val == q->val)&&isEqual(p->left,q->left) && isEqual(p->right,q->right);    }

1003. 相连的1

循环查找到1则进行深度搜索，查找到1则置为0

void changeRightBottom(vector<vector<char>>& A, int x, int y, int& r, int& c) {    A[x][y] = '0';    if(x+1 < r && A[x+1][y]=='1') changeRightBottom(A,x+1,y,r,c);    if(x > 0 && A[x-1][y]=='1')changeRightBottom(A,x-1,y,r,c);    if(y+1 < c && A[x][y+1]=='1') changeRightBottom(A,x,y+1,r,c);    if(y > 0 && A[x][y-1]=='1') changeRightBottom(A,x,y-1,r,c);}int countConnectedOnes(vector<vector<char>>& A) {    int row = A.size();    if(row == 0) return 0;    int col = A[0].size();    if(col == 0) return 0;    int res = 0;    for(int i = 0; i < row; i++) {        for(int j = 0; j < col; j++) {            if(A[i][j]!='0') {                changeRightBottom(A, i, j, row, col);                res++;            }        }    }    return res;}

1004.判断无环图

转换成图后，进行深度优先搜索，有重复入栈为有环图，返回false

bool DFS(vector<vector<int>>& g, vector<int>& s, int index){    if(find(s.begin(), s.end(), index) != s.end())        return false;    s.push_back(index);    for(int i = 0; i < g[index].size(); i++) {        if(!DFS(g, s, g[index][i]))            return false;    }    s.pop_back();    return true;}bool isDAG(int n, vector<pair<int, int>>& edges) {    if(n < 2) return true;    vector<vector<int>> graph(n);    for(int i = 0; i < edges.size(); i++) {        int in = edges[i].first, out = edges[i].second;        graph[in].push_back(out);    }    vector<int> s;    for(int i = 0; i < n; i++){        if(!DFS(graph, s, i))return false;    }    return true;}

1005.最大和

动态规划：起始状态为a[0] = A[0], a[1] = max(A[1], A[0]), 转移方程a[i] = max(A[i] + a[i-2], a[i-1])

    int maxSum(vector<int>& A) {          int a[100000] = {0};          a[0] = A[0];          a[1] = A[1] > A[0] ? A[1]:A[0];          for(int i = 2; i < A.size(); i++){              a[i] = max(A[i] + a[i-2], a[i-1]);          }           return a[A.size() - 1];    }

1006.单词变换

课本原题：dp[i][j]表示长度为i的word1与长度为j的word2

int minDistance(string word1, string word2) {    int l1 = word1.size(),        l2 = word2.size(),         ll1 = l1+1,         ll2 = l2+1;    vector<vector<int>> d(ll1, vector<int>(ll2, 0));    for(int i = 0; i < ll2; i++)        d[0][i] = i;    for(int i = 0; i < ll1; i++)        d[i][0] = i;    for(int i = 1; i < ll1; i++) {        for(int j = 1; j < ll2; j++) {            if(word1[i-1] == word2[j-1])                d[i][j] = d[i-1][j-1];            else {                int m = min(d[i-1][j], d[i][j-1]);                d[i][j] = min(m, d[i-1][j-1]) + 1;            }        }    }    return d[l1][l2];}

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