LeetCode题解：Spiral Matrix I and II

Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]]

You should return [1,2,3,6,9,8,7,4,5].

Spiral Matrix II

Given an integer n, generate a square matrix filled with elements from 1 ton² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ]]

思路：

光标转移有四个状态，依次是向右，向下，向左，向上。光标转移状态的变化根据三个条件：1. 到达上界（n）2.到达下界（0）3.光标的下一个位置已经被填满。据此编码。

题解：

class Solution {public:    vector<int> spiralOrder(vector<vector<int> > &matrix) {        enum { VISITED = -1 };                  if (matrix.size()==0)            return vector<int>();                    const int M = matrix.size();        const int N = matrix[0].size();                int cursorX = 0;          int cursorY = 0;                    int incrX = 1;          int incrY = 0;                    int nextX, nextY;                  int LEFT = 0, RIGHT = N;        int TOP = 0, BOTTOM = M;                  auto valid_X=[&](const int c) -> bool {              return c >= LEFT && c < RIGHT;          };                  auto valid_Y=[&](const int c) -> bool {            return c >= TOP && c < BOTTOM ;        };                  auto valid_state=[&]()->bool {              return valid_X(nextX) && valid_Y(nextY);        };                    auto next_state=[&]() {              switch(incrX * 2 + incrY)            {                case 2: // moving right                    ++TOP;                    incrX = 0, incrY = 1;                    break;                case -2: // moving left                    --BOTTOM;                    incrX = 0, incrY = -1;                    break;                case 1: // moving down                    --RIGHT;                    incrX = -1, incrY = 0;                    break;                case -1: // moving up                    ++LEFT;                    incrX = 1, incrY = 0;                    break;            }        };                    auto next_pos=[&]() {              nextX = cursorX + incrX;              nextY = cursorY + incrY;          };                vector<int> ret;        ret.reserve(M * N);        for(int i = 0; i < M * N; ++i)        {            ret.push_back(matrix[cursorY][cursorX]);            next_pos();            if (!valid_state())              {                  next_state();                  next_pos();              }                         cursorX = nextX;            cursorY = nextY;        }                return ret;    }};

class Solution {public:    vector<vector<int> > generateMatrix(int n) {        enum { UNUSED = -1 };                vector<vector<int>> matrix(n);        for(auto& row : matrix)        {            row.resize(n);            fill(begin(row), end(row), UNUSED);        }                int cursorX = 0;        int cursorY = 0;                int incrX = 0;        int incrY = 1;                int nextX, nextY;                auto valid_pos=[&](const int c) -> bool {            return c >= 0 && c < n;        };                auto valid_state=[&]()->bool {            return valid_pos(nextX) &&                 valid_pos(nextY) &&                matrix[nextX][nextY] == UNUSED;        };                auto next_state=[&]() {            switch(incrX * 2 + incrY)            {                case 1:                    incrX = 1, incrY = 0;                    break;                case 2:                    incrX = 0, incrY = -1;                    break;                case -1:                    incrX = -1, incrY = 0;                    break;                case -2:                    incrX = 0, incrY = 1;                    break;            }        };                auto next_pos=[&]() {            nextX = cursorX + incrX;            nextY = cursorY + incrY;        };                for(int i = 1; i <= n * n; ++i)        {            matrix[cursorX][cursorY] = i;                        // increment the cursor            next_pos();            if (!valid_state())            {                next_state();                next_pos();            }                        cursorX = nextX;            cursorY = nextY;        }                return matrix;    }};