本文介绍了2个不同二维数组的总和的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!
问题描述
我需要帮助编写一个方法来找出 2 个不同大小的二维数组的总和.
I need help writing a method to find out a sum of 2 different sized 2D arrays.
public static int[][] summary(int[][] tab1, int[][] tab2, int x) {
int[][] finalTab = new int[4][5]; // I took sizes of bigger one
if (x < 0) {
for (int i = 0; i < finalTab.length - 1; i++) {
for (int j = 0; j < finalTab[i].length - 1; j++) {
finalTab[i][j] = tab1[i][j] + tab2[i][j];
if (tab1[i][j] == 0) {
finalTab[i][j] = tab2[i][j];
}
}
}
for (int i = 0; i < finalTab.length; i++) {
for (int j = 0; j < finalTab[i].length; j++) {
System.out.print(" " + finalTab[i][j] + " ");
}
System.out.println();
}
}
return finalTab;
}
输入是:
int[][] tab1 = {
{2, 4, 6, 8},
{2, 4, 6, 8},
{2, 4, 6, 8}};
int[][] tab2 = {
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9},
{1, 3, 5, 7, 9}};
输出是:
3 7 11 15 0
3 7 11 15 0
3 7 11 15 0
0 0 0 0 0
输出应该是:
3 7 11 15 9
3 7 11 15 9
3 7 11 15 9
1 3 5 7 9
如何将0
全部替换为tab2
的索引对应的数字?
How can I replace all of 0
with numbers corresponding to the index of tab2
?
推荐答案
你可以做得更通用
public static int[][] summary(int[][] tab1, int[][] tab2, int x) {
int maxLenX = tab1.length > tab2.length ? tab1.length : tab2.length;
int maxLenY = tab1[0].length > tab2[0].length ? tab1[0].length : tab2[0].length;
int[][] finalTab = new int[maxLenX][maxLenY]; // i took sizes of bigger one
if (x < 0) {
for (int i = 0; i <= finalTab.length - 1; i++) {
for (int j = 0; j <= finalTab[i].length - 1; j++) {
if (i > tab1.length - 1 || j > tab1[i].length - 1) {
finalTab[i][j] = tab2[i][j];
} else if (i > tab2.length - 1 || j > tab2[i].length - 1) {
finalTab[i][j] = tab1[i][j];
} else {
finalTab[i][j] = tab1[i][j] + tab2[i][j];
}
}
}
for (int i = 0; i < finalTab.length; i++) {
for (int j = 0; j < finalTab[i].length; j++) {
System.out.print(" " + finalTab[i][j] + " ");
}
System.out.println();
}
}
return finalTab;
}
所以你可以这样称呼它
summary(tab2, tab1, -1);
或
summary(tab1, tab2, -1);
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