Spiral matrix output arrays in spiral order

[Copy question]:

Given a matrix of m x n elements (m rows, n columns), return all elements of the Matri X in Spiral Order.

Example 1:

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

Example 2:

input:[  [1, 2, 3, 4],  [5, 6, 7, 8],  [9,10,11,12]]output: [1,2,3,4,8,12,11,10,9,5,6,7]

[Brute force solution]:

Time Analysis:

Spatial Analysis:

[After optimization]:

Time Analysis:

Spatial Analysis:

[Wonderful output CONDITIONS]:

[Wonderful corner case]:

After the variable is changed, the if control range is used, otherwise it will be out of bounds:

if <= rowend && colbegin <= colend)

[Thinking questions]:

The feeling indicates that the variable of corner is always changed, bad expression. New four new variables on the line, anyway do not occupy space.

[English data structure or algorithm, why not other data structures or algorithms]:

[a sentence of thought]:

[input]: null: Normal: Large: Extra Small: Special cases handled in the program: abnormal conditions (unreasonable input):

[Drawing]:

[One brush]:

1. Start/end are to be added ( including in) the number, so be sure to write the equal sign

[Two brushes]:

[Three brushes]:

[Four brushes]:

[Five brushes]:

[Results of five-minute visual debug]:

[Summary]:

Newly opened several recorded variables that do not occupy space

[Complexity]:time Complexity:o (n) Space complexity:o (n)

[Algorithmic thinking: Iterative/recursive/split/greedy]:

[Key templating code]:

[Other solutions]:

[Follow up]:

[The problem given by the LC becomes variable]:

[Code Style]:

[Whether to write this type of driver Funcion code for the first time]:

[Subtext]:

 Public classSolution { PublicList<integer> Spiralorder (int[] matrix) {List<Integer> res =NewArraylist<integer>(); if(Matrix.length = = 0) {            returnRes; }                intRowbegin = 0; intRowend = matrix.length-1; intColbegin = 0; intColend = matrix[0].length-1;  while(Rowbegin <= rowend && colbegin <=colend) {            //Traverse Right             for(intj = Colbegin; J <= Colend; J + +) {Res.add (matrix[rowbegin][j]); } rowbegin++; //Traverse Down             for(intj = Rowbegin; J <= Rowend; J + +) {Res.add (matrix[j][colend]); } colend--; if(Rowbegin <= rowend && colbegin <=colend) {                //Traverse Left                 for(intj = colend; J >= Colbegin; J--) {Res.add (matrix[rowend][j]); }} rowend--; if(Colbegin <= colend && rowbegin <=rowend) {                //Traver up                 for(intj = rowend; J >= Rowbegin; J--) {Res.add (Matrix[j][colbegin]); }} Colbegin++; }                returnRes; }}

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Spiral matrix output arrays in spiral order