tags: DSA C++ LeetCode Python
写在前面
螺旋矩阵系列, 严格来说不算双指针, 但是其中蕴含的思想很像双指针. (应该叫四指针)
- 54. 螺旋矩阵 - 力扣（LeetCode）;(需要四个指针分别在需要转弯的时候移动)
- 59. 螺旋矩阵 II - 力扣（LeetCode）;(跟上面的题异曲同工)
- 885. 螺旋矩阵 III - 力扣（LeetCode）;(不需要考虑边界直接模拟, 注意这个题是从内往外转, 需要定义方向数组)
- 2326. 螺旋矩阵 IV - 力扣（LeetCode）;(同基本的螺旋矩阵, 加上链表向后遍历的基本操作即可)
螺旋矩阵I
只能说, 用Python不讲武德:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]: res = [] while matrix: # 削头（第一层） res += matrix.pop(0) # 将剩下的逆时针转九十度，等待下次被削 matrix = list(zip(*matrix))[::-1] return res
C++的写法很简练, 思路直接在代码中体现出来了. 四个变量逐次更新.
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size(), SIZE = m * n;
int l{}, r{n - 1}, t{}, b{m - 1}, i{}, x{}, y{};
vector<int> ans(SIZE);
while (i < SIZE) {
while (y <= r && i < SIZE) ans[i++] = matrix[t][y++];
++t, x = t;
while (x <= b && i < SIZE) ans[i++] = matrix[x++][r];
--r, y = r;
while (y >= l && i < SIZE) ans[i++] = matrix[b][y--];
--b, x = b;
while (x >= t && i < SIZE) ans[i++] = matrix[x--][l];
++l, y = l;
}
return ans;
}
};
// 用for循环也一样: 可能看起来简练一些
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size(), SIZE = m * n;
int l{}, r{n - 1}, t{}, b{m - 1}, i{}, x{}, y{};
vector<int> ans(SIZE);
while (i < SIZE) {
for (y = l; y <= r && i < SIZE; ++y) ans[i++] = matrix[t][y];
++t;
for (x = t; x <= b && i < SIZE; ++x) ans[i++] = matrix[x][r];
--r;
for (y = r; y >= l && i < SIZE; --y) ans[i++] = matrix[b][y];
--b;
for (x = b; x >= t && i < SIZE; --x) ans[i++] = matrix[x][l];
++l;
}
return ans;
}
};或者用一种定义方向数组的写法, 算是一种模板了.
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size(), SIZE = m * n;
// 方向: 右下左上
int dirs[4][2] {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int k{}, r{}, c{}, d{}, i{}, j{};
vector<int> ans(SIZE);
while (k < SIZE) {
ans[k++] = matrix[i][j];
matrix[i][j] = 101; // 标记遍历过
r = i + dirs[d][0], c = j + dirs[d][1];
// 换向
if (r < 0 || r >= m || c < 0 || c >= n || matrix[r][c] == 101)
d = (d + 1) % 4, r = i + dirs[d][0], c = j + dirs[d][1];
i = r, j = c;
}
return ans;
}
};螺旋矩阵II
第一题代码改改还能用:
class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
int SIZE = n * n;
int l{}, r{n - 1}, t{}, b{n - 1}, i{1}, x{}, y{};
vector<vector<int>> ans(n, vector<int>(n));
while (i <= SIZE) {
for (y = l; y <= r; ++y) ans[t][y] = i++;
++t;
for (x = t; x <= b; ++x) ans[x][r] = i++;
--r;
for (y = r; y >= l; --y) ans[b][y] = i++;
--b;
for (x = b; x >= t; --x) ans[x][l] = i++;
++l;
}
return ans;
}
};定义方向数组的方法: (不容易想, 但是代码相对简洁)
class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
int SIZE = n * n, k{1};
// 方向: 右下左上
int dirs[4][2]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int r{}, c{}, d{}, i{}, j{};
vector<vector<int>> ans(n, vector<int>(n));
while (k <= SIZE) {
ans[i][j] = k++;
r = i + dirs[d][0], c = j + dirs[d][1];
// 换向
if (r < 0 || r >= n || c < 0 || c >= n || ans[r][c])
d = (d + 1) % 4, r = i + dirs[d][0], c = j + dirs[d][1];
i = r, j = c;
}
return ans;
}
};螺旋矩阵III
这个题和1,2,4不太一样, 原因在于旋转是由内而外的了, 这就要考虑一下边界情况了.
用定义四个方向数组的方法套模板就可以.
class Solution {
public:
vector<vector<int>> spiralMatrixIII(int rows, int cols, int rStart,
int cStart) {
int dirs[4][2]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // 顺时针, 右下左上
int SIZE = rows * cols, i{1}, steps{}, r{rStart}, c{cStart},
d{};
vector<vector<int>> ans(rows * cols);
ans[0] = {r, c};
while (i < SIZE) {
++steps;
// 每旋转两个方向, 就要增加一次单向距离
for (int p{}; p < 2; ++p) {
for (int j{}; j < steps; ++j) {
auto& [rx, cx] = dirs[d]; // C++17
r += rx, c += cx;
if (r >= 0 && r < rows && c >= 0 && c < cols)
ans[i++] = {r, c};
}
d = (d + 1) % 4;
}
}
return ans;
}
};螺旋矩阵IV
熟悉一下链表的遍历, 这道题就会了. 直接用第一题代码.
class Solution {
public:
vector<vector<int>> spiralMatrix(int m, int n, ListNode* head) {
vector<vector<int>> ans(m, vector<int>(n, -1));
int t{}, l{}, r{n - 1}, b{m - 1}, i{}, j{};
while (head) {
for (j = l; j <= r && head; ++j, head = head->next)
ans[t][j] = head->val;
++t;
for (i = t; i <= b && head; ++i, head = head->next)
ans[i][r] = head->val;
--r;
for (j = r; j >= l && head; --j, head = head->next)
ans[b][j] = head->val;
--b;
for (i = b; i >= t && head; --i, head = head->next)
ans[i][l] = head->val;
++l;
}
return ans;
}
};方向数组版:
class Solution {
public:
vector<vector<int>> spiralMatrix(int m, int n, ListNode* head) {
// 方向: 右下左上
int dirs[4][2]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int r{}, c{}, d{}, i{}, j{};
vector<vector<int>> ans(m, vector<int>(n, -1));
while (head) {
ans[i][j] = head->val, head = head->next;
r = i + dirs[d][0], c = j + dirs[d][1];
// 换向
if (r < 0 || r >= m || c < 0 || c >= n || ans[r][c] != -1)
d = (d + 1) % 4, r = i + dirs[d][0], c = j + dirs[d][1];
i = r, j = c;
}
return ans;
}
};通用方法
可以看出熟悉套路之后I,II,IV都可以迎刃而解, 但是III需要考虑的多一些,
class Solution {
public:
vector<vector<int>> spiralMatrix(int m, int n, ListNode* head) {
// 方向数组: 右下左上
int dirs[4][2]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
// 循环变量
int r{}, c{}, d{}, i{}, j{};
// 初始化结果数组
vector<int> ans(m * n);
while (/* 满足循环条件 */) {
// 更新结果
ans[i][j] = k
r = i + dirs[d][0], c = j + dirs[d][1];
if (r < 0 || r >= m || c < 0 || c >= n || /* 元素被遍历过? */)
// 换方向, 更新步
d = (d + 1) % 4, r = i + dirs[d][0], c = j + dirs[d][1];
// 更新索引
i = r, j = c;
}
return ans;
}
};
