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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.
1 class Solution { 2 public: 3 4 static bool cmp(int a,int b) 5 { 6 int digitA=1; 7 int digitB=1; 8 9 int tmpA=a; 10 11 while(tmpA>=10) 12 { 13 digitA++; 14 tmpA/=10; 15 } 16 17 int tmpB=b; 18 while(tmpB>=10) 19 { 20 digitB++; 21 tmpB/=10; 22 } 23 24 return a*pow(10,digitB)+b>b*pow(10,digitA)+a; 25 26 } 27 28 string largestNumber(vector<int> &num) { 29 30 sort(num.begin(),num.end(),cmp); 31 32 string result=""; 33 34 for(int i=0;i<num.size();i++) 35 { 36 result+=to_string(num[i]); 37 } 38 39 if(result[0]==‘0‘) result="0"; 40 41 return result; 42 43 } 44 };
另一种方法,先把数字转化为字符串,然后通过字符串拼接去比较,这样cmp函数会好些很多
1 class Solution { 2 public: 3 4 static bool cmp(string a,string b) 5 { 6 return a+b>b+a; 7 } 8 9 string largestNumber(vector<int> &num) { 10 11 int n=num.size(); 12 vector<string> strNum(n); 13 for(int i=0;i<n;i++) 14 { 15 strNum[i]=to_string(num[i]); 16 } 17 18 sort(strNum.begin(),strNum.end(),cmp); 19 20 string result=""; 21 22 for(int i=0;i<n;i++) 23 { 24 result+=strNum[i]; 25 } 26 27 if(result[0]==‘0‘) result="0"; 28 29 return result; 30 } 31 };
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原文地址:http://www.cnblogs.com/reachteam/p/4222486.html