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!\n");
            exit(0);
        }
    }
    printf("No cycles found.\n");
    return 0;
}

 

posted @ 2020-05-10 09:34  繁星的夜空2012  阅读(368)  评论(0编辑  收藏  举报