Largest Number

Given a list of non negative integers, arrange them such that they form the largest number.

Have you met this question in a real interview? Yes
Example
Given [1, 20, 23, 4, 8], the largest formed number is 8423201.

Note
The result may be very large, so you need to return a string instead of an integer.

public class Solution {  
    public String largestNumber(int[] num) {  
        int n = num.length;  
        if (n < 1) return "";  

        //把数组转化为字符串数组  
        String[] strs = new String[n];  
        for (int i = 0; i < n; i++) {  
            strs[i] = String.valueOf(num[i]);  
        }  

        //按照数字在结果中的位置排序  
        Arrays.sort(strs, new Cmp());  

        //把所有字符串拼接在一起  
        String ans = "";  
        for (int i = n - 1; i >= 0; i--) {  
            ans = ans.concat(strs[i]);  
        }  

        //去掉数字开头的0,如输入[0, 0]  
        int i = 0;  
        while (i < n && ans.charAt(i) == '0') {  
            i++;  
        }  
        if (i == n) return "0";  

        return ans.substring(i);  
    }  
}  

//比较方法很简单:如果[a在前b在后]组成的结果大于[b在前a在后]组成的结果,那么a>b。  
class Cmp implements Comparator<String>{  
    @Override  
    public int compare(String a, String b) {  
       String   ab = a.concat(b);
        String  ba = b.concat(a);

        return (ab).compareTo(ba);
    }  
} 
这段代码有问题,修改一下,MOV r0, #0x00002000 ; Initialize pointer to first number MOV r1, #9 ; Initialize counter with number of elements LDR r7, [r0] ; Load first number as largest LDR r8, [r0] ; Load first number as smallest Loop: ADD r0, r0, #4 ; Move pointer to next number LDR r2, [r0] ; Load the number in r2 CMP r7, r2 ; Compare largest with current number MOVLT r7, r2 ; If current number is smaller, update largest CMP r8, r2 ; Compare smallest with current number MOVGT r8, r2 ; If current number is larger, update smallest SUBS r1, r1, #1 ; Decrement counter BNE Loop ; Loop until all numbers are compared ; Display largest number on console MOV r0, #1 ; File descriptor for stdout LDR r1, =largest ; Address of string to be displayed MOV r2, #10 ; Length of string MOV r7, #4 ; Syscall number for write SWI 0 ; Call operating system ; Display largest number on LCD screen LDR r0, =0x40020C14 ; Address of LCD data register MOV r1, r7 ; Load largest number from r7 STR r1, [r0] ; Store the number in the LCD data register ; Display smallest number on console MOV r0, #1 ; File descriptor for stdout LDR r1, =smallest ; Address of string to be displayed MOV r2, #12 ; Length of string MOV r7, #4 ; Syscall number for write SWI 0 ; Call operating system ; Display smallest number on LCD screen LDR r0, =0x40020C14 ; Address of LCD data register MOV r1, r8 ; Load smallest number from r8 STR r1, [r0] ; Store the number in the LCD data register largest: .asciz "Largest number: %d\n" smallest: .asciz "Smallest number: %d\n"
05-27
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值