54. 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].
难度:87,这道题跟 Rotate Image 很相似,都是需要把矩阵分割成层来处理,每一层都是按:1. 正上方;2. 正右方;3. 正下方;4. 正左方这种顺序添加元素到结果集合。实现中要注意细节,when I traverse left or up I have to check whether the row or col still exists to prevent duplicates.
public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<Integer>();
if (matrix.length == 0) {
return res;
}
int rowBegin = 0;
int rowEnd = matrix.length-1;
int colBegin = 0;
int colEnd = matrix[0].length - 1;
while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse Right
for (int j = colBegin; j <= colEnd; j ++) {
res.add(matrix[rowBegin][j]);
}
rowBegin++;
// Traverse Down
for (int j = rowBegin; j <= rowEnd; j ++) {
res.add(matrix[j][colEnd]);
}
colEnd--;
if (rowBegin <= rowEnd) {
// Traverse Left
for (int j = colEnd; j >= colBegin; j --) {
res.add(matrix[rowEnd][j]);
}
}
rowEnd--;
if (colBegin <= colEnd) {
// Traver Up
for (int j = rowEnd; j >= rowBegin; j --) {
res.add(matrix[j][colBegin]);
}
}
colBegin ++;
}
return res;
}
}
矩阵非图的题就得找规律, 自己在一遍遍的走, 然后看看遍历顺序和开始行结束行, 开始列结束列的关系, 自己找一找corner case, 看看是否有越界问题.
然后就是得分情况讨论, 记住coner case

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