問題描述

我需要幫助編寫一個方法來找出 2 個不同大小的二維數(shù)組的總和.

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;
}

輸入是:

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}};

輸出是:

 3  7  11  15  0 
 3  7  11  15  0 
 3  7  11  15  0 
 0  0  0  0  0 

輸出應(yīng)該是:

 3  7  11  15  9
 3  7  11  15  9
 3  7  11  15  9
 1  3  5  7  9

如何將0全部替換為tab2的索引對應(yīng)的數(shù)字?

How can I replace all of 0 with numbers corresponding to the index of tab2?

推薦答案

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;
}

所以你可以這樣稱呼它

summary(tab2, tab1, -1);

或

summary(tab1, tab2, -1);

這篇關(guān)于2個不同二維數(shù)組的總和的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！