【题目】给定一个矩阵matrix,按照“之”字形的方式打印这个矩阵,例如:
1,8,6,7
2,6,4,11
3,5,9,10
打印结果是1,8,2,3,6,6,7,4,5,9,11,10。要求额外空间复杂度是O(1)。
思路:
A逐个向右移动,B逐个向下移动,如果当两个标记有一个超过了界限,B则向右移动,A向下移动,两个坐标始终处于一条斜线,然后打印斜线上的元素就行。
/**
“之字形打印矩阵”
*/
public class Code_08_ZigZagPrintMatrix {
public static void printMatrixZigZag(int[][] matrix) {
int tR = 0;
int tC = 0;
int dR = 0;
int dC = 0;
int endR = matrix.length - 1;
int endC = matrix[0].length - 1;
boolean fromUp = false;
while (tR != endR + 1) {
printLevel(matrix, tR, tC, dR, dC, fromUp);
tR = tC == endC ? tR + 1 : tR;
tC = tC == endC ? tC : tC + 1;
dC = dR == endR ? dC + 1 : dC;
dR = dR == endR ? dR : dR + 1;
fromUp = !fromUp;
}
System.out.println();
}
public static void printLevel(int[][] m, int tR, int tC, int dR, int dC,
boolean f) {
if (f) {
while (tR != dR + 1) {
System.out.print(m[tR++][tC--] + " ");
}
} else {
while (dR != tR - 1) {
System.out.print(m[dR--][dC++] + " ");
}
}
}
public static void main(String[] args) {
int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } };
printMatrixZigZag(matrix);
}
}