算法模块总结

Posted 2020-12-15 maleyang

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