给定一个包含 m x n 个元素的矩阵(m 行, n 列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。
示例 1:
输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
解法1:
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> list = new ArrayList<>();
if(matrix == null || matrix.length == 0) return list;
int start = 0;
while(matrix[0].length > start * 2 && matrix.length > start * 2){
printOneCircle(matrix, start, list);
start++;
}
return list;
}
public void printOneCircle(int[][] matrix, int start, List<Integer> list){
int endX = matrix[0].length - 1 - start;//列
int endY = matrix.length - 1 - start;//行
//从左往右
for(int i = start; i <= endX; i++){
list.add(matrix[start][i]);
}
//从上到下
if(start < endY){
for(int i = start + 1; i <= endY; i++){
list.add(matrix[i][endX]);
}
}
//从右到左
if(start < endX && start < endY){
for(int i = endX - 1; i >= start; i--){
list.add(matrix[endY][i]);
}
}
//从下到上
if(start < endX && start < endY - 1){
for(int i = endY - 1; i >= start + 1; i--){
list.add(matrix[i][start]);
}
}
}
解法2:
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> res = new ArrayList<>();
if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
return res;
}
int tR = 0; //左上角的行
int tC = 0; //左上角的列
int dR = matrix.length - 1; //右下角的行
int dC = matrix[0].length - 1; //右下角的列
while(tR <= dR && tC <= dC){
printEdge(matrix, tR++, tC++, dR--, dC--, res);
}
return res;
}
public void printEdge(int[][] matrix, int tR, int tC, int dR, int dC, List<Integer> res){
if(tR == dR){ //左下角的行与右下角的行相等
for(int i = tC; i <= dC; i++){
res.add(matrix[tR][i]);
}
}else if(tC == dC){//左下角的列与右下角的列相等
for(int i = tR; i <= dR; i++){
res.add(matrix[i][tC]);
}
}else{
int curC = tC;
int curR = tR;
while(curC != dC){
res.add(matrix[tR][curC]);
curC++;
}
while(curR != dR){
res.add(matrix[curR][dC]);
curR++;
}
while(curC != tC){
res.add(matrix[dR][curC]);
curC--;
}
while(curR != tR){
res.add(matrix[curR][tC]);
curR--;
}
}
}