• 算法模块总结


    层次遍历算法总结

    C语言的层次遍历总结篇
    1、先定义一个队列的结构体
    typedef struct {
        int x;
        int y;
    } Node;
    
    int numIslands(char** grid, int gridSize, int* gridColSize){
        2、鲁棒性、判断输入参数
        if(grid == NULL || gridSize <= 0 || gridColSize[0] <= 0){
            return 0;
        }
        3、根据输入参数、返回值的需要定义
        int landsNums = 0;
        4、模拟队列,队列初始化
        Node* queue = (Node*)malloc(sizeof(Node)*gridSize*gridColSize[0]);
        memset(queue, 0, sizeof(Node)*gridSize*gridColSize[0]);
        5、存储临时变量作为当前值
        Node cur;
        6、根据当前选择列表进行层次遍历算法
        for(int i = 0;  i < gridSize; i++){
            for(int j = 0; j < gridColSize[0]; j++){
                if(grid[i][j] == '1'){
                    
                    grid[i][j] = '0';
                    7、对于选择列表的每个节点都进行层次遍历,则需要头指针和尾指针初始化
                    int front = 0;
                    queue[front].x = i;
                    queue[front++].y = j;
                    8、取出当前队列的节点,并将其子节点入队列
                    while(front != 0){
                        cur = queue[--front];
                        if(cur.x - 1 >= 0 && grid[cur.x-1][cur.y] == '1'){
                            grid[cur.x-1][cur.y] = '0';
                            queue[front].x = cur.x-1;
                            queue[front++].y = cur.y;
                        }
                        if (cur.x + 1 < gridSize && grid[cur.x + 1][cur.y] == '1') {
                            grid[cur.x + 1][cur.y] = '0';
                            queue[front].x = cur.x + 1;
                            queue[front++].y = cur.y;
                        }
                        if (cur.y - 1 >= 0 && grid[cur.x][cur.y - 1] == '1') {
                            grid[cur.x][cur.y - 1] = '0';
                            queue[front].x = cur.x;
                            queue[front++].y = cur.y - 1;
                        }
                        if (cur.y + 1 < gridColSize[0] && grid[cur.x][cur.y + 1] == '1') {
                            grid[cur.x][cur.y + 1] = '0';
                            queue[front].x = cur.x;
                            queue[front++].y = cur.y + 1;
                        }
                    }
                    9、当前题目的需求进行处理
                    landsNums++;
                }
            }
        }
        free(queue);
        return landsNums;
        
    }

     被围绕的区域

    130. 被围绕的区域
    https://leetcode-cn.com/problems/surrounded-regions/
    typedef struct {
        int x;
        int y;
    } Node;
    void solve(char** board, int boardSize, int* boardColSize){
        if(board == NULL || boardSize <= 0 || boardColSize[0] <= 0){
            return ;
        }
        
        Node* queue = (Node*)malloc(sizeof(Node)*boardSize*boardColSize[0]);
        memset(queue, 0, sizeof(Node)*boardSize*boardColSize[0]);
        Node cur;
        
        for(int i = 0; i < boardSize; i++){
            for(int j = 0; j < boardColSize[0]; j++){
                if((i==0 || i == boardSize-1 || j == 0 || j ==boardColSize[0]-1) && board[i][j] == 'O'){
                    int front = 0;
                    int rear = 0;
                    queue[front].x = i;
                    queue[front++].y = j;
                    while(front != 0){
                        cur = queue[--front];
                        if(cur.x - 1 >= 0 && board[cur.x-1][cur.y] == 'O'){
                            board[cur.x-1][cur.y] = 'm';
                            queue[front].x = cur.x-1;
                            queue[front++].y = cur.y;
                        }
                        if (cur.x + 1 < boardSize && board[cur.x + 1][cur.y] == 'O') {
                            board[cur.x + 1][cur.y] = 'm';
                            queue[front].x = cur.x + 1;
                            queue[front++].y = cur.y;
                        }
                        if (cur.y - 1 >= 0 && board[cur.x][cur.y - 1] == 'O') {
                            board[cur.x][cur.y - 1] = 'm';
                            queue[front].x = cur.x;
                            queue[front++].y = cur.y - 1;
                        }
                        if (cur.y + 1 < boardColSize[0] && board[cur.x][cur.y + 1] == 'O') {
                            board[cur.x][cur.y + 1] = 'm';
                            queue[front].x = cur.x;
                            queue[front++].y = cur.y + 1;
                        }
                        
                    }
                    
                    board[i][j] = 'm';
                    
                }
            }
        }
            // 遍历矩阵,把O全部改写成X,A全部改写成O
        for (int i = 0; i < boardSize; i++) {
            for (int j = 0; j < boardColSize[0]; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                } else if (board[i][j] == 'm') {
                    board[i][j] = 'O';
                }
            }
        }
        return;
    
    }
    130. 被围绕的区域

    并查集算法

    并查集算法
    #include <stdio.h>
    #include <stdlib.h>
    #define VERTICES 6
     
    void initialise(int parent[], int rank[]){
        int i;
        for(i = 0; i < VERTICES; i++){
            parent[i] = -1;
            rank[i] = 0;
        }
    }
    
    int find_root(int x, int parent[]){
        int x_root = x;
        while(parent[x_root] != -1){
            x_root = parent[x_root];
        }
        return x_root;
    }
    
    int union_vertices(int x, int y, int parent[], int rank[]){
        int x_root = find_root(x, parent);
        int y_root = find_root(y, parent);
        if(x_root == y_root){
            return 0;
        } else {
            if(rank[x_root] > rank[y_root]){
                parent[y_root] = x_root;
            }else if(rank[y_root] > rank[x_root]){
                parent[y_root] = x_root;
            }else {
                parent[x_root] = y_root;
                rank[y_root]++;
            }
            return 1;
        }
    }
    
    int main(){
        int parent[VERTICES] = {0};
        int rank[VERTICES] = {0};
        int edges[5][2] = {
            {0,1}, {1,2}, {1,3},
            {3,4}, {2,5}
        };
        initialise(parent,rank);
        int i;
        for(i = 0; i < 5; i++){
            int x = edges[i][0];
            int y = edges[i][1];
            if(union_vertices(x,y,parent,rank) == 0){
                printf("Cycle detected!
    ");
                exit(0);
            }
        }
        printf("No cycles found.
    ");
        return 0;
    }

    在代码的世界尽情的翱翔吧!
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  • 原文地址:https://www.cnblogs.com/maleyang/p/12862218.html
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