Largest Number

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

For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.

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

Credits:

Special thanks to @ts for adding this problem and creating all test cases.


先考虑字符串的长短

class Solution {
    string itoa(int num) {
        stringstream stream;
        stream << num;
        return stream.str();
    }
    
    static bool comp(const string& str1, const string& str2) {
        if(str1.size() > str2.size() || str2.empty()) return !comp(str2, str1);
        
        if(str1.empty()) {
            return !str2.empty();
        }
        
        int i = 0;
        while( i < str1.size() ) {
            if(str1[i] > str2[i]) return true;
            if(str1[i] < str2[i]) return false;
            i++;
        }
        string str2_next = i== str2.size() ? "" : str2.substr(i);
        return comp(str1, str2_next);
        return false;
    }
    
public:
    string largestNumber(vector<int>& nums) {
        vector<string> rst;
        for( int num : nums ) {
            rst.push_back(itoa(num));
        }
        sort(rst.begin(), rst.end(), comp);
        
        string str;
        for( string s : rst ) {
            str.append(s);
        }
        int i = 0;
        while( i < str.size() && str[i] == '0') i++;
        return i == str.size() ? "0" : str.substr(i);
    }
};


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