2023-05-11：給你一個 m x n 的二進(jìn)制矩陣 grid，
每個格子要么為 0 （空）要么為 1 （被占據(jù)），
給你郵票的尺寸為 stampHeight x stampWidth。
我們想將郵票貼進(jìn)二進(jìn)制矩陣中，且滿足以下 限制 和 要求 ：
覆蓋所有空格子，不覆蓋任何被占據(jù)的格子，
可以放入任意數(shù)目的郵票，郵票可以相互有重疊部分，
郵票不允許旋轉(zhuǎn)，郵票必須完全在矩陣內(nèi)，
如果在滿足上述要求的前提下，可以放入郵票，請返回 true ，否則返回 false。
輸入：grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3。
輸出：true。
答案2023-05-11：
大體過程如下：
1.首先對矩陣 grid 進(jìn)行二維前綴和計算，得到一個新的矩陣 sum。該矩陣中每個位置表示從左上角出發(fā)，到該位置形成的子矩陣中所有元素的和。
2.對 grid 中的每個為 0 的位置 (i, j)，檢查以該位置為左上角的子矩陣是否能夠被指定的印章完全覆蓋。如果可以，將 diff[i][j] 加 1，diff[i][j+stampWidth] 減 1，diff[i+stampHeight][j] 減 1，diff[i+stampHeight][j+stampWidth] 加 1。這里 diff 矩陣用于記錄每個位置的變化量。
3.遍歷 grid 中的每一行，使用滾動數(shù)組的方式還原 cnt 和 pre 數(shù)組，并通過它們來計算每列中為 0 的位置的數(shù)量。同時，如果某個位置 (i, j) 的值為 0 且它所在列中沒有其他的 0，則返回 false；否則返回 true。
時間復(fù)雜度為 O(mn)，其中 m 和 n 分別表示矩陣 grid 的行數(shù)和列數(shù)。這是因為函數(shù)需要遍歷整個矩陣，并對每個位置進(jìn)行常數(shù)次操作。同時，二維前綴和、二維差分和滾動數(shù)組優(yōu)化的時間復(fù)雜度也都是 O(mn)。
空間復(fù)雜度為 O(mn)，因為函數(shù)中創(chuàng)建了兩個 m+1 行 n+1 列的二維數(shù)組 sum 和 diff，以及一個長度為 n+1 的一維數(shù)組 cnt 和 pre。這些數(shù)組所占用的總空間為 (m+1)(n+1) + 2(n+1) = mn + 3m + 3n + 3，即 O(mn)。
go完整代碼如下：
package main
import "fmt"
func main() {
grid := [][]int{{1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}}
stampHeight := 4
stampWidth := 3
isPossibleToStamp := possibleToStamp(grid, stampHeight, stampWidth)
fmt.Println(isPossibleToStamp)
}
func possibleToStamp(grid [][]int, stampHeight, stampWidth int) bool {
m, n := len(grid), len(grid[0])
sum := make([][]int, m+1)
sum[0] = make([]int, n+1)
diff := make([][]int, m+1)
diff[0] = make([]int, n+1)
for i, row := range grid {
sum[i+1] = make([]int, n+1)
for j, v := range row { // grid 的二維前綴和
sum[i+1][j+1] = sum[i+1][j] + sum[i][j+1] - sum[i][j] + v
}
diff[i+1] = make([]int, n+1)
}
for i, row := range grid {
for j, v := range row {
if v == 0 {
x, y := i+stampHeight, j+stampWidth // 注意這是矩形右下角橫縱坐標(biāo)都 +1 后的位置
if x <= m && y <= n && sum[x][y]-sum[x][j]-sum[i][y]+sum[i][j] == 0 {
diff[i][j]++
diff[i][y]--
diff[x][j]--
diff[x][y]++ // 更新二維差分
}
}
}
}
// 還原二維差分矩陣對應(yīng)的計數(shù)矩陣，這里用滾動數(shù)組實現(xiàn)
cnt := make([]int, n+1)
pre := make([]int, n+1)
for i, row := range grid {
for j, v := range row {
cnt[j+1] = cnt[j] + pre[j+1] - pre[j] + diff[i][j]
if cnt[j+1] == 0 && v == 0 {
return false
}
}
cnt, pre = pre, cnt
}
return true
}
rust完整代碼如下：
fn main() {
let grid = vec![
vec![1, 0, 0, 0],
vec![1, 0, 0, 0],
vec![1, 0, 0, 0],
vec![1, 0, 0, 0],
vec![1, 0, 0, 0],
];
let stamp_height = 4;
let stamp_width = 3;
let is_possible_to_stamp = possible_to_stamp(&grid, stamp_height, stamp_width);
println!("{}", is_possible_to_stamp);
}
fn possible_to_stamp(grid: &[Vec<i32>], stamp_height: usize, stamp_width: usize) -> bool {
let m = grid.len();
let n = grid[0].len();
let mut sum = vec![vec![0; n + 1]; m + 1];
let mut diff = vec![vec![0; n + 1]; m + 1];
for i in 0..m {
for j in 0..n {
sum[i + 1][j + 1] = sum[i + 1][j] + sum[i][j + 1] - sum[i][j] + grid[i][j];
}
}
for i in 0..m {
for j in 0..n {
if grid[i][j] == 0 {
let x = i + stamp_height;
let y = j + stamp_width;
if x <= m && y <= n && sum[x][y] - sum[x][j] - sum[i][y] + sum[i][j] == 0 {
diff[i][j] += 1;
diff[i][y] -= 1;
diff[x][j] -= 1;
diff[x][y] += 1;
}
}
}
}
let mut cnt = vec![0; n + 1];
let mut pre = vec![0; n + 1];
for i in 0..m {
for j in 0..n {
cnt[j + 1] = cnt[j] + pre[j + 1] - pre[j] + diff[i][j];
if cnt[j + 1] == 0 && grid[i][j] == 0 {
return false;
}
}
std::mem::swap(&mut cnt, &mut pre);
}
true
}
c語言完整代碼如下：
#include <stdio.h>
#include <stdlib.h>
int possibleToStamp(int** grid, int gridSize, int* gridColSize, int stampHeight, int stampWidth) {
int m = gridSize, n = *gridColSize;
int** sum = (int**)malloc(sizeof(int*) * (m + 1));
for (int i = 0; i <= m; i++) {
sum[i] = (int*)malloc(sizeof(int) * (n + 1));
}
int** diff = (int**)malloc(sizeof(int*) * (m + 1));
for (int i = 0; i <= m; i++) {
diff[i] = (int*)malloc(sizeof(int) * (n + 1));
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sum[i + 1][j + 1] = sum[i + 1][j] + sum[i][j + 1] - sum[i][j] + grid[i][j];
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0) {
int x = i + stampHeight, y = j + stampWidth;
if (x <= m && y <= n && sum[x][y] - sum[x][j] - sum[i][y] + sum[i][j] == 0) {
diff[i][j]++;
diff[i][y]--;
diff[x][j]--;
diff[x][y]++;
}
}
}
}
int* cnt = (int*)malloc(sizeof(int) * (n + 1));
int* pre = (int*)malloc(sizeof(int) * (n + 1));
for (int i = 0; i <= n; i++) {
cnt[i] = 0;
pre[i] = 0;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
cnt[j + 1] = cnt[j] + pre[j + 1] - pre[j] + diff[i][j];
if (cnt[j + 1] == 0 && grid[i][j] == 0) {
free(cnt);
free(pre);
for (int k = 0; k <= m; k++) {
free(sum[k]);
}
free(sum);
for (int k = 0; k <= m; k++) {
free(diff[k]);
}
free(diff);
return 0;
}
}
int* tmp = cnt;
cnt = pre;
pre = tmp;
}
free(cnt);
free(pre);
for (int i = 0; i <= m; i++) {
free(sum[i]);
}
free(sum);
for (int i = 0; i <= m; i++) {
free(diff[i]);
}
free(diff);
return 1;
}
int main() {
int gridSize = 5, gridColSize = 4;
int** grid = (int**)malloc(sizeof(int*) * gridSize);
for (int i = 0; i < gridSize; i++) {
grid[i] = (int*)malloc(sizeof(int) * gridColSize);
}
int data[5][4] = { {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0} };
for (int i = 0; i < gridSize; i++) {
for (int j = 0; j < gridColSize; j++) {
grid[i][j] = data[i][j];
}
}
int stampHeight = 4, stampWidth = 3;
int isPossibleToStamp = possibleToStamp(grid, gridSize, &gridColSize, stampHeight, stampWidth);
printf("%s\n", isPossibleToStamp ? "true" : "false");
for (int i = 0; i < gridSize; i++) {
free(grid[i]);
}
free(grid);
return 0;
}
文章來源：http://www.zghlxwxcb.cn/news/detail-699554.html
c++完整代碼如下：
#include <iostream>
#include <vector>
using namespace std;
bool possibleToStamp(vector<vector<int>>& grid, int stampHeight, int stampWidth) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> sum(m + 1, vector<int>(n + 1)), diff(m + 1, vector<int>(n + 1));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sum[i + 1][j + 1] = sum[i + 1][j] + sum[i][j + 1] - sum[i][j] + grid[i][j];
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0) {
int x = i + stampHeight, y = j + stampWidth;
if (x <= m && y <= n && sum[x][y] - sum[x][j] - sum[i][y] + sum[i][j] == 0) {
diff[i][j]++;
diff[i][y]--;
diff[x][j]--;
diff[x][y]++;
}
}
}
}
vector<int> cnt(n + 1), pre(n + 1);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
cnt[j + 1] = cnt[j] + pre[j + 1] - pre[j] + diff[i][j];
if (cnt[j + 1] == 0 && grid[i][j] == 0) {
return false;
}
}
swap(cnt, pre);
}
return true;
}
int main() {
vector<vector<int>> grid{ {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0}, {1, 0, 0, 0} };
int stampHeight = 4, stampWidth = 3;
bool isPossibleToStamp = possibleToStamp(grid, stampHeight, stampWidth);
cout << (isPossibleToStamp ? "true" : "false") << endl;
return 0;
}
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到了這里，關(guān)于2023-05-11：給你一個 m x n 的二進(jìn)制矩陣 grid， 每個格子要么為 0 （空）要么為 1 （被占據(jù)）， 給你郵票的尺寸為 stampHeight x stampWidth。 我們想將的文章就介紹完了。如果您還想了解更多內(nèi)容，請在右上角搜索TOY模板網(wǎng)以前的文章或繼續(xù)瀏覽下面的相關(guān)文章，希望大家以后多多支持TOY模板網(wǎng)！
