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English Version

题目描述

给定 下标从 0 开始 m x n 二进制 矩阵 grid

在一次操作中,可以选择满足以下条件的任意 ij:

  • 0 <= i < m
  • 0 <= j < n
  • grid[i][j] == 1

并将第 i 行和第 j 列中的 所有 单元格的值更改为零。

返回从 grid 中删除所有 1 所需的最小操作数。

 

示例 1:

输入: grid = [[1,1,1],[1,1,1],[0,1,0]]
输出: 2
解释:
在第一个操作中,将第 1 行和第 1 列的所有单元格值更改为 0。
在第二个操作中,将第 0 行和第 0 列的所有单元格值更改为 0。

示例 2:

输入: grid = [[0,1,0],[1,0,1],[0,1,0]]
输出: 2
解释:
在第一个操作中,将第 1 行和第 0 列的所有单元格值更改为 0。
在第二个操作中,将第 2 行和第 1 列的所有单元格值更改为 0。
注意,我们不能使用行 1 和列 1 执行操作,因为 grid[1][1]!= 1。

示例 3:

输入: grid = [[0,0],[0,0]]
输出: 0
解释:
没有 1 可以移除,所以返回0。

 

提示:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 15
  • 1 <= m * n <= 15
  • grid[i][j] 为 0 或 1

解法

方法一:状态压缩 + BFS

Python3

class Solution:
    def removeOnes(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if grid[i][j])
        q = deque([state])
        vis = {state}
        ans = 0
        while q:
            for _ in range(len(q)):
                state = q.popleft()
                if state == 0:
                    return ans
                for i in range(m):
                    for j in range(n):
                        if grid[i][j] == 0:
                            continue
                        nxt = state
                        for r in range(m):
                            nxt &= ~(1 << (r * n + j))
                        for c in range(n):
                            nxt &= ~(1 << (i * n + c))
                        if nxt not in vis:
                            vis.add(nxt)
                            q.append(nxt)
            ans += 1
        return -1

Java

class Solution {
    public int removeOnes(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int state = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    state |= 1 << (i * n + j);
                }
            }
        }
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(state);
        Set<Integer> vis = new HashSet<>();
        vis.add(state);
        int ans = 0;
        while (!q.isEmpty()) {
            for (int k = q.size(); k > 0; --k) {
                state = q.poll();
                if (state == 0) {
                    return ans;
                }
                for (int i = 0; i < m; ++i) {
                    for (int j = 0; j < n; ++j) {
                        if (grid[i][j] == 0) {
                            continue;
                        }
                        int nxt = state;
                        for (int r = 0; r < m; ++r) {
                            nxt &= ~(1 << (r * n + j));
                        }
                        for (int c = 0; c < n; ++c) {
                            nxt &= ~(1 << (i * n + c));
                        }
                        if (!vis.contains(nxt)) {
                            vis.add(nxt);
                            q.offer(nxt);
                        }
                    }
                }
            }
            ++ans;
        }
        return -1;
    }
}

C++

class Solution {
public:
    int removeOnes(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int state = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (grid[i][j])
                    state |= (1 << (i * n + j));
        queue<int> q {{state}};
        unordered_set<int> vis {{state}};
        int ans = 0;
        while (!q.empty()) {
            for (int k = q.size(); k > 0; --k) {
                state = q.front();
                q.pop();
                if (state == 0) return ans;
                for (int i = 0; i < m; ++i) {
                    for (int j = 0; j < n; ++j) {
                        if (grid[i][j] == 0) continue;
                        int nxt = state;
                        for (int r = 0; r < m; ++r) nxt &= ~(1 << (r * n + j));
                        for (int c = 0; c < n; ++c) nxt &= ~(1 << (i * n + c));
                        if (!vis.count(nxt)) {
                            vis.insert(nxt);
                            q.push(nxt);
                        }
                    }
                }
            }
            ++ans;
        }
        return -1;
    }
};

Go

func removeOnes(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    state := 0
    for i, row := range grid {
        for j, v := range row {
            if v == 1 {
                state |= 1 << (i * n + j)
            }
        }
    }
    q := []int{state}
    vis := map[int]bool{state:true}
    ans := 0
    for len(q) > 0 {
        for k := len(q); k > 0; k-- {
            state = q[0]
            if state == 0 {
                return ans
            }
            q = q[1:]
            for i, row := range grid {
                for j, v := range row {
                    if v == 0 {
                        continue
                    }
                    nxt := state
                    for r := 0; r < m; r++ {
                        nxt &= ^(1 << (r * n + j))
                    }
                    for c := 0; c < n; c++ {
                        nxt &= ^(1 << (i * n + c))
                    }
                    if !vis[nxt] {
                        vis[nxt] = true
                        q = append(q, nxt)
                    }
                }
            }
        }
        ans++
    }
    return -1
}

TypeScript

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