LeetCode054 Spiral Matrix

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详细见：leetcode.com/problems/spiral-matrix

Java Solution: github

package leetcode;import java.util.LinkedList;import java.util.List;public class P054_SpiralMatrix {public static void main(String[] args) {System.out.println(new Solution().spiralOrder(new int[][]{{1},{1},{1},{1}}));}/* * 最重要的是代码简洁，否则就是没用 * 1 ms * 2.73% */static class Solution {List<Integer> ans = new LinkedList<Integer>();    public List<Integer> spiralOrder(int[][] matrix) {    int row = 0, col = 0, eni = 0;    if (matrix == null || (row = matrix.length) == 0     || (col = matrix[0].length) == 0)    return ans;    eni = Math.min(row, col) >> 1;    for (int i = 0; i != eni; i ++)    fourEdge(i, row - 1 - i, col - 1 - i, matrix);    if ((Math.min(row, col) & 0x1) == 1) {    int rowi = Math.min(row, col) >> 1;    int staj = rowi;    int endj = Math.max(row, col) - rowi - 1;    if (row < col)    for (int j = staj; j <= endj; j ++)    ans.add(matrix[rowi][j]);    else    for (int i = staj; i <= endj; i ++)    ans.add(matrix[i][rowi]);    }    return ans;    }private void fourEdge(int i, int j, int k, int[][] matrix) {int t = 0;for (t = i; t < k; t ++)ans.add(matrix[i][t]);for (t = i; t < j; t ++)ans.add(matrix[t][k]);for (t = k; t > i; t --)ans.add(matrix[j][t]);for (t = j; t > i; t --)ans.add(matrix[t][i]);}}}

C Solution: github

/*    url: leetcode.com/problems/spiral-matrix/    AC 3ms 0.00%*/#include <stdio.h>#include <stdlib.h>void shell_add(int* ans, int* ai, int** m, int x1, int y1, int x2, int y2) {    int x = 0, y = 0;    for (y = y1; y < y2; y ++)        ans[(*ai) ++] = m[x1][y];    for (x = x1; x < x2; x ++)        ans[(*ai) ++] = m[x][y2];    for (y = y2; y > y1; y --)        ans[(*ai) ++] = m[x2][y];    for (x = x2; x > x1; x --)        ans[(*ai) ++] = m[x][y1];}int* spiralOrder(int** m, int rn, int cn) {    int* ans = (int*) malloc(sizeof(int)*(rn*cn));    int ai = 0;    int x1 = 0, y1 = 0, x2 = rn - 1, y2 = cn - 1;    while (x1 < x2 && y1 < y2) {        shell_add(ans, &ai, m, x1, y1, x2, y2);        x1 ++;        y1 ++;        x2 --;        y2 --;    }    if (x1 == x2) {        while(y1 <= y2)            ans[ai ++] = m[x1][y1++];    } else if (y1 == y2) {        while (x1 <= x2)            ans[ai ++] = m[x1++][y1];    }    printf("ai is %d\r\n", ai);    return ans;}int main() {    int rn = 7, cn = 3;    int** m = (int**) malloc(sizeof(int*)*rn);    int m0[] = {1, 2, 3};    int m1[] = {4, 5, 6};    int m2[] = {7, 8, 9};    int m3[] = {10, 11, 12};    int m4[] = {13, 14, 15};    int m5[] = {16, 17, 18};    int m6[] = {19, 20, 21};    int* ans = NULL;    int i = 0;    m[0] = m0;    m[1] = m1;    m[2] = m2;    m[3] = m3;    m[4] = m4;    m[5] = m5;    m[6] = m6;    ans = spiralOrder(m, rn, cn);    for (i = 0; i < rn*cn; i ++)        printf("%d ", ans[i]);    printf("\r\n");    free(m);}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/spiral-matrix    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月10日    @details:    Solution: 45ms 46.70%'''class Solution(object):    def shell_add(self, a, ai, i1, j1, i2, j2, m):        for k in range(j1, j2):            a[ai[0]] = m[i1][k]            ai[0] += 1        for k in range(i1 , i2):            a[ai[0]] = m[k][j2]            ai[0] += 1        for k in range(j2, j1, -1):            a[ai[0]] = m[i2][k]            ai[0] += 1        for k in range(i2, i1, -1):            a[ai[0]] = m[k][j1]            ai[0] += 1            def spiralOrder(self, m):        """        :type m: List[List[int]]        :rtype: List[int]        """        mi = 0 if m == None else len(m)        if mi == 0: return []        mj = 0 if m[0] == None else len(m[0])        if mj == 0: return []                i1, j1, i2, j2, ai, a = 0,0,mi-1,mj-1,[0],[0]*(mi*mj)        while i1 < i2 and j1 < j2:            self.shell_add(a, ai, i1, j1, i2, j2, m)            i1, i2, j1, j2 = i1+1,i2-1,j1+1,j2-1        if i1 == i2:            for k in range(j1, j2+1):                a[ai[0]] = m[i1][k]                ai[0] += 1        elif j1 == j2:            for k in range(i1, i2+1):                a[ai[0]] = m[k][j1]                ai[0] += 1        return  a

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