共计 3894 个字符，预计需要花费 10 分钟才能阅读完成。
No.1 能够造成最大正方形的矩形数目
解题思路
保护最大正方形的边长并计数即可。
代码展现
class Solution {public int countGoodRectangles(int[][] rectangles) {
int maxLen = 0, count = 0;
for (int[] rec : rectangles) {int len = Math.min(rec[0], rec[1]);
if (len > maxLen) {
maxLen = len;
count = 0;
}
if (len == maxLen) {count++;}
}
return count;
}
}No.2 同积元组
解题思路
用 Map 记录每一种乘积的因数列表。
代码展现
class Solution {public int tupleSameProduct(int[] nums) {Arrays.sort(nums);
// map[i] 示意乘积为 i 的较小的因数的列表
// 比方 a * b == i 则 map[i].add(min(a, b))
Map<Integer, List<Integer>> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {for (int j = i + 1; j < nums.length; j++) {int a = nums[i], b = nums[j];
int p = a * b;
if (!map.containsKey(p)) {map.put(p, new ArrayList<>());
}
map.get(p).add(a); // a < b
}
}
// 有 len 对乘积为 i
// 它们能形成 4 * len * (len - 1) 个元组
int count = 0;
for (var entry : map.entrySet()) {int len = entry.getValue().size();
count += 4 * len * (len - 1);
}
return count;
}
}No.3 重新排列后的最大子矩阵
解题思路
枚举最大子矩阵的起始行，而后排序、统计即可。
代码展现
class Solution {public int largestSubmatrix(int[][] matrix) {Column[] cols = new Column[matrix[0].length];
for (int i = 0; i < matrix[0].length; i++) {int[] col = new int[matrix.length];
for (int j = 0; j < matrix.length; j++) {col[j] = matrix[j][i];
}
cols[i] = new Column(col);
}
int res = 0;
// 枚举最大子矩阵的起始行 i, 而后排序
for (int i = 0; i < matrix.length; i++) {
final int idx = i;
Arrays.sort(cols, (a, b) -> b.cont1[idx] - a.cont1[idx]);
if (res < cols[0].cont1[idx] * cols.length) { // 一个优化点
res = Math.max(res, count(cols, idx));
}
}
return res;
}
// 统计以后场面下的最大子矩阵
private int count(Column[] cols, int startRow) {
int res = 0;
for (int endRow = startRow, endCol = cols.length - 1; endRow < cols[0].raw.length; endRow++) {while (endCol >= 0 && cols[endCol].raw[endRow] == 0) {endCol--;}
res = Math.max(res, (endRow - startRow + 1) * (endCol + 1));
}
return res;
}
// Column 类示意一列
// raw 示意这一列的原始值
// cont1[i] 示意 raw[i] 后有多少个间断的 1
class Column {int[] raw;
int[] cont1;
Column(int[] col) {
raw = col;
cont1 = new int[col.length];
cont1[col.length - 1] = col[col.length - 1];
for (int i = col.length - 2; i >= 0; i--) {cont1[i] = col[i];
if (col[i] == 1) {cont1[i] += cont1[i + 1];
}
}
}
}
}No.4 猫和老鼠 II
解题思路
动静布局（记忆化搜寻）
尽管代码量比拟大，然而逻辑并不简单。实质上依然是枚举猫和老鼠的下一步怎么走，老鼠想要赢就要保障它走下一步之后，猫无论怎么走都赢不了。而后应用记忆化以提速。
代码展现
class Solution {
int n, m;
char[][] grid;
int[][][][][] catMemo; // 记忆化
int[][][][][] mouseMemo; // 记忆化
int catJump, mouseJump;
final int[][] dirs = new int[][] {{ -1, 0}, {1, 0}, {0, 1}, {0, -1} };
final int MaxRound = 100; // 8x8 的地图远不须要 1000 步
public boolean valid(int i, int j) {return i >= 0 && i < n && j >= 0 && j < m && grid[i][j] != '#';
}
// 曾经走了 r 步，老鼠在 (mx, my) 猫在 (cx, cy)
// 猫是否必胜
public int cat(int r, int mx, int my, int cx, int cy) {if (catMemo[r][mx][my][cx][cy] == -1) {if (mx == cx && my == cy) {return catMemo[r][mx][my][cx][cy] = 1;
}
if (grid[mx][my] == 'F') {return catMemo[r][mx][my][cx][cy] = 0;
}
if (grid[cx][cy] == 'F') {return catMemo[r][mx][my][cx][cy] = 1;
}
for (int[] d : dirs) {for (int jump = 0; jump <= catJump; jump++) {int x = cx + d[0] * jump;
int y = cy + d[1] * jump;
if (!valid(x, y)) {break;}
if (mouse(r + 1, mx, my, x, y) == 0) {return catMemo[r][mx][my][cx][cy] = 1;
}
}
}
catMemo[r][mx][my][cx][cy] = 0;
}
return catMemo[r][mx][my][cx][cy];
}
// 曾经走了 r 步，老鼠在 (mx, my) 猫在 (cx, cy)
// 老鼠是否必胜
public int mouse(int r, int mx, int my, int cx, int cy) {if (r >= MaxRound) {return 0;}
if (mouseMemo[r][mx][my][cx][cy] == -1) {if (mx == cx && my == cy) {return mouseMemo[r][mx][my][cx][cy] = 0;
}
if (grid[mx][my] == 'F') {return mouseMemo[r][mx][my][cx][cy] = 1;
}
if (grid[cx][cy] == 'F') {return mouseMemo[r][mx][my][cx][cy] = 0;
}
for (int[] d : dirs) {for (int jump = 0; jump <= mouseJump; jump++) {int x = mx + d[0] * jump;
int y = my + d[1] * jump;
if (!valid(x, y)) {break;}
if (cat(r, x, y, cx, cy) == 0) {return mouseMemo[r][mx][my][cx][cy] = 1;
}
}
}
mouseMemo[r][mx][my][cx][cy] = 0;
}
return mouseMemo[r][mx][my][cx][cy];
}
public boolean canMouseWin(String[] grid, int catJump, int mouseJump) {
n = grid.length;
m = grid[0].length();
this.grid = new char[n][m];
int mx = 0, my = 0, cx = 0, cy = 0;
for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {this.grid[i][j] = grid[i].charAt(j);
if (this.grid[i][j] == 'C') {
cx = i;
cy = j;
}
if (this.grid[i][j] == 'M') {
mx = i;
my = j;
}
}
}
this.catJump = catJump;
this.mouseJump = mouseJump;
catMemo = new int[MaxRound][n][m][n][m];
mouseMemo = new int[MaxRound][n][m][n][m];
// 初值 -1 示意未计算过
for (int r = 0; r < MaxRound; r++) {for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {for (int t = 0; t < n; t++) {for (int k = 0; k < m; k++) {catMemo[r][i][j][t][k] = -1;
mouseMemo[r][i][j][t][k] = -1;
}
}
}
}
}
return mouse(0, mx, my, cx, cy) == 1;
}
}
正文完
