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].
Idea: spiral array, need to control the direction of output, my implementation is to define a Boolean array, the same size as the array, the data is initially all false. Then define an int variable to represent the direction, the data boundary or the next data is true, then change direction until it crosses or is all true. Ends the loop.
The specific code is as follows:
public class Solution {public list<integer> Spiralorder (int[][] a) {list<integer> List = new ArrayList <Integer> (); if (A.length = = 0 | | a[0].length = = 0) {return list; } int i = 0;//line int j = 0;//column boolean[][] b = new Boolean[a.length][a[0].length]; int o = 0;//for direction, 0: Right, 1: Lower, 2: Left, 3: On//within range, out of range over while (I < a.length && I >= 0 &&am P J < A[0].length && J >= 0) {if (B[i][j]) {//If all has gone, end loop break; } list.add (A[i][j]); Add result b[i][j] = true;//has been added as true to indicate that the switch (o) {case 0://is in the right direction. if (j = = A[0].length-1 | | b[i][j+1]) {o = 1;//go to the right or marked, go down in the direction i++; }else{J + +; } break; Case 1: if (i = = A.length-1 | | b[i+1][j]) {o = 2;//go to the bottom or marked, direction left to go j--; }else{i++; } break; Case 2:if (j = = 0 | | b[i][j-1]) {o = 3;//go to leftmost or marked, direction upward i--; }else{j--; } break; Case 3:if (i = = 0 | | b[i-1][j]) {o = 0;//go to the top or marked, direction to the right. j + +; }else{i--; } break; }} return list; }}
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Leetcode 54.Spiral matrix (spiral matrix) thinking and method of solving problems