标签：初始 题目 span 入队 返回 tar problem free pen
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; }
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; }
标签：初始 题目 span 入队 返回 tar problem free pen
原文地址：https://www.cnblogs.com/maleyang/p/12862218.html
