矩陣題目

1. 輸出對角線上半個(gè)矩陣都是1的矩陣

image.png

上半個(gè)矩陣是1,通過i < j的條件可以取到所有1的點(diǎn),其他都是0

public class Solution {
    public static void main(String args[] ) throws Exception {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt(); // read an integer from input stream.
        // System.out.println(n);  // print n
        outputMatrix(n);
    }

    public static void outputMatrix(int n){
        int[][] m = new int[n][n];
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < n; i++){
            sb = new StringBuilder();
            for(int j = 0; j < n;j++){
                if(i < j){
                    m[i][j] = 1;
                }
                else{
                    m[i][j] = 0;
                }
                sb.append(m[i][j]+ " ");
            }
            System.out.println(sb.toString().trim());
        }
    }
}

2.輸出沙漏型hourglass矩陣

沙漏中是1,沙漏外是0


image.png

硬著數(shù),
每一行分成三份,n是奇數(shù)odd number or偶數(shù) even number要分情況討論,上半個(gè)矩陣和下半個(gè)矩陣是相反的情況討論

public class Solution {
    public static void main(String args[] ) throws Exception {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt(); // read an integer from input stream.
        // System.out.println(n);  // print n
        printoutMatrix(n);
    }

    public static void printoutMatrix(int n){
        int[][] m = new int[n][n];
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < n; i++){
            sb = new StringBuilder();
            if(n % 2 == 0) { // n is even number
                if (i < n / 2) { // i row number is upper half
                    for (int j = 0; j < i; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = i; j <= n - 1 - i; j++) {
                        m[i][j] = 1;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = n - 1 - i + 1; j <= n - 1; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                } else { // i row number is downside half
                    for (int j = 0; j < n - 1 - i; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = n - 1 - i; j <= i; j++) {
                        m[i][j] = 1;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = i + 1; j <= n - 1; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                }
            }
            else if(n % 2 == 1) { //n is odd number
                if (i <= n / 2) { //row number i is upper half
                    for (int j = 0; j < i; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = i; j <= n - 1 - i; j++) {
                        m[i][j] = 1;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = n - 1 - i + 1; j <= n - 1; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                } else { //row number is downside half
                    for (int j = 0; j < n - 1 - i; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = n - 1 - i; j <= i; j++) {
                        m[i][j] = 1;
                        sb.append(m[i][j] + " ");
                    }
                    for (int j = i + 1; j <= n - 1; j++) {
                        m[i][j] = 0;
                        sb.append(m[i][j] + " ");
                    }
                }
            }
           
            System.out.println(sb.toString().trim());
            
        }
    }
}
?著作權(quán)歸作者所有,轉(zhuǎn)載或內(nèi)容合作請聯(lián)系作者
【社區(qū)內(nèi)容提示】社區(qū)部分內(nèi)容疑似由AI輔助生成,瀏覽時(shí)請結(jié)合常識與多方信息審慎甄別。
平臺聲明:文章內(nèi)容(如有圖片或視頻亦包括在內(nèi))由作者上傳并發(fā)布,文章內(nèi)容僅代表作者本人觀點(diǎn),簡書系信息發(fā)布平臺,僅提供信息存儲服務(wù)。

相關(guān)閱讀更多精彩內(nèi)容

友情鏈接更多精彩內(nèi)容