317. Shortest Distance from All Buildings

题目：

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

Example 1:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

链接： http://leetcode.com/problems/maximum-product-of-word-lengths/

题解：

给一块空地，求在空地上新盖一座楼，到其他楼的距离最短。这里我们要用到BFS,就是从每个楼开始用BFS计算空地 "0"到这栋楼的距离，最后把每个空地到每栋楼的距离加起来，求一个最小值。这里我们还要算出楼的数目，仅当空地能连接所有楼的时候，我们才愿意在这块空地上造楼。我们也要维护一个visited矩阵来避免重复。

Time Complexity - O(4^mn)， Space Complexity - O(mn * k)， k为1的数目

public class Solution {
    private final int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    
    public int shortestDistance(int[][] grid) {
        if(grid == null || grid.length == 0) {
            return Integer.MAX_VALUE;
        }
        int rowNum = grid.length;
        int colNum = grid[0].length;
        int[][] distance = new int[rowNum][colNum];
        int[][] canReachBuildings = new int[rowNum][colNum];
        int buildingNum = 0; 
        
        for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(grid[i][j] != 0) {
                    distance[i][j] = Integer.MAX_VALUE;
                }
                if(grid[i][j] == 1) {     // find out all buildings
                    buildingNum++;
                    updateDistance(grid, distance, canReachBuildings, i, j);
                }
            }
        }
        
        int min = Integer.MAX_VALUE;
        for(int i = 0; i < rowNum; i++) {
            for(int j = 0; j < colNum; j++) {
                if(canReachBuildings[i][j] == buildingNum) {
                    min = Math.min(distance[i][j], min);
                }
            }
        }
        
        return min == Integer.MAX_VALUE ? -1 : min;
    }
    
    private void updateDistance(int[][] grid, int[][] distance, int[][] canReachBuildings, int row, int col) {
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{row, col});
        boolean[][] visited = new boolean[grid.length][grid[0].length];
        visited[row][col] = true;
        int dist = 0;
        int curLevel = 1;
        int nextLevel = 0;
        
        while(!queue.isEmpty()) {
            int[] position = queue.poll();
            distance[position[0]][position[1]] += dist;
            curLevel--;
            for(int[] direction : directions) {
                int x = position[0] + direction[0];
                int y = position[1] + direction[1];
                if(x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != 0) {
                    continue;
                }
                if(!visited[x][y]) {
                    queue.offer(new int[]{x, y}); 
                    nextLevel++;
                    visited[x][y] = true;
                    canReachBuildings[x][y]++;
                }
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                dist++;
            }
        }
    }
}

 

Reference:

https://leetcode.com/discuss/74453/36-ms-c-solution

https://leetcode.com/discuss/74422/clean-solution-easy-understanding-with-simple-explanation

https://leetcode.com/discuss/74999/java-solution-with-explanation-and-time-complexity-analysis

https://leetcode.com/discuss/74380/my-bfs-java-solution