[LeetCode] 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.

解题思路：先按照一定的规则对数字进行排序，排序规则是：对两个变量从最高位开始进行逐位比较，当只有其中一个变量还有未比较位数时，从该位数字开始，与另一变量进行逐位比较。当该变量只有一位时，则看该位是否不小于另一变量中的最大数字。

实现代码：

/*****************************************************************************
    *  @COPYRIGHT NOTICE
    *  @Copyright (c) 2015, 楚兴
    *  @All rights reserved
    *  @Version  : 1.0

    *  @Author   : 楚兴
    *  @Date     : 2015/2/6 13:23
    *  @Status   : Accepted
    *  @Runtime  : 9 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
	string largestNumber(vector<int> &num) {
		Solution s;
		string str = "";
		sort(num.begin(),num.end(),s);
		char ch[11];
		int sum = 0;
		for (auto it = num.begin(); it != num.end(); it++)
		{
			sum += *it;
			itoa(*it,ch,10);
			str += ch;
		}
		return sum == 0 ? "0" : str;
	}

	bool operator()(int m, int n)
	{
		if (m == n)
		{
			return false;
		}
		char ch1[11];
		char ch2[12];
		itoa(m,ch1,10);
		itoa(n,ch2,10);
		char* t1 = ch1;
		char* t2 = ch2;
		while (*t1 && *t2)
		{
			if (*t1 != *t2)
			{
				return *t1 > *t2;
			}
			t1++;
			t2++;
		}
		while (*t1)
		{
			return strcmp(t1,ch2);
		}
		while(*t2)
		{
			return !strcmp(t2,ch1);
		}
	}
private:
	char maxchar(char* ch)
	{
		char* temp = ch;
		char max = '\0';
		while(*temp)
		{
			max = max > *temp ? max : *temp;
			temp++;
		}
		return max;
	}
    //如果ch1应在前，返回true，否则返回false
	bool strcmp(char* ch1, char* ch2)
	{
		char* t1 = ch1;
		char* t2 = ch2; // ch2位数较短
		while (*t1)
		{
			if (*t2 == '\0')
			{
				t2 = ch2; //指向首字符
			}
			if (*t1 != *t2)
			{
				return *t1 > *t2; 
			}
			else
			{
				if (*(t1 + 1) == '\0')
				{
					char c = maxchar(ch2);
					bool b = *t1 >= c;
					return b;
				}
			}
			t1++;
			t2++;
		}
		return false;
	}

	void itoa(int n, char ch[11], int R = 10)
	{
		char* temp = ch;
		if (n == 0)
		{
			strcpy(ch,"0");
			return;
		}
		while(n)
		{
			*temp++ = n % R + '0';
			n = n / R;
		}
		*temp = '\0';
		reverse(ch);
	}

	void reverse(char ch[11])
	{
		char* p = ch;
		char* q = ch + strlen(ch) - 1;
		char c;
		while(p < q)
		{
			c = *p;
			*p++ = *q;
			*q-- = c;
		}
	}
};
int main()
{
	vector<int> nums;
	nums.push_back(8308);
	nums.push_back(830);
	for (int i = 9; i >= 0; i--)
	{
		nums.push_back(i);
	}
	nums.push_back(2281);
	nums.push_back(121);
	nums.push_back(12);

	Solution s;
	string str = s.largestNumber(nums);
	cout<<str.c_str()<<endl;
	system("pause");
}

第二种思路：省去前面找排序思路的复杂过程，直接先排序，再检验。

/*****************************************************************************
    *  @COPYRIGHT NOTICE
    *  @Copyright (c) 2015, 楚兴
    *  @All rights reserved
    *  @Version  : 2.0

    *  @Author   : 楚兴
    *  @Date     : 2015/2/6 14:06
    *  @Status   : Accepted
    *  @Runtime  : 12 ms
*****************************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
	string largestNumber(vector<int> &num) {
		Solution s;
		string str = "";
		sort(num.begin(),num.end(),s);
		char ch[11];
		int sum = 0;
		for (auto it = num.begin(); it != num.end(); it++)
		{
			sum += *it;
			itoa(*it,ch,10);
			str += ch;
		}
		return sum == 0 ? "0" : str;
	}

	bool operator()(int m, int n)
	{
		if (m == 0) //等于0时不能放前面
		{
			return false;
		}
		
		char ch1[11];
		char ch2[11];
		itoa(m,ch1);
		itoa(n,ch2);
		if (strlen(ch1) == strlen(ch2))
		{
			return m >= n;
		}
		
		char ch3[22];
		char ch4[22];
		memcpy(ch3,ch1,sizeof(ch1));
		strcat(ch3,ch2);
		memcpy(ch4,ch2,sizeof(ch2));
		strcat(ch4,ch1);

		char* t1 = ch3;
		char* t2 = ch4;
		while(*t1)
		{
			if (*t1 != *t2)
			{
				return *t1 > *t2;
			}
			t1++;
			t2++;
		}
		return true;
	}
private:
	void strcat(char* input, char* output)
	{
		char* t1 = input;
		char* t2 = output;
		while (*t1)
		{
			t1++;
		}
		while(*t2)
		{
			*t1++ = *t2++;
		}
		*t1 = '\0';
	}

	void itoa(int n, char ch[11], int R = 10)
	{
		char* temp = ch;
		if (n == 0)
		{
			strcpy(ch,"0");
			return;
		}
		while(n)
		{
			*temp++ = n % R + '0';
			n = n / R;
		}
		*temp = '\0';
		reverse(ch);
	}

	void reverse(char ch[11])
	{
		char* p = ch;
		char* q = ch + strlen(ch) - 1;
		char c;
		while(p < q)
		{
			c = *p;
			*p++ = *q;
			*q-- = c;
		}
	}
};
int main()
{
	vector<int> nums;
	nums.push_back(8308);
	nums.push_back(830);
	for (int i = 9; i >= 0; i--)
	{
	nums.push_back(0);
	}
	nums.push_back(2281);
	nums.push_back(121);
	nums.push_back(12);

	Solution s;
	string str = s.largestNumber(nums);
	cout<<str.c_str()<<endl;
	char ch[10];
	memset(ch,0,2);
	cout<<strlen(ch)<<endl;
	system("pause");
}