标签：大小 char public 去掉 compare ges integer compareto rri
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.
聪明方法：其实干嘛要挨个比呢，按最直接的方法，接起来，谁大谁在前：
可以换一下思路，要想比较两个数在最终结果中的先后位置，何不直接比较一下不同组合的结果大小？
举个例子：要比较3和34的先后位置，可以比较334和343的大小，而343比334大，所以34应当在前。
这样，有了比较两个数的方法，就可以对整个数组进行排序。然后再把排好序的数拼接在一起就好了。
首先把int 全部转换成string array，然后，自己写一个comparator，判断ab ba的大小，从而把a，b排序
然后把所有的连起来，记住，大的在后面，从后面开始连接。最后去掉前面的0；
public class Solution {
public String largestNumber(int[] num) {
if(num == null || num.length == 0)
return "";
// Convert int array to String array, so we can sort later on
String[] s_num = new String[num.length];
for(int i = 0; i < num.length; i++)
s_num[i] = String.valueOf(num[i]);
// Comparator to decide which string should come first in concatenation
Comparator<String> comp = new Comparator<String>(){
@Override
public int compare(String str1, String str2){
String s1 = str1 + str2;
String s2 = str2 + str1;
return s2.compareTo(s1); // reverse order here, so we can do append() later
}
};
Arrays.sort(s_num, comp);
// An extreme edge case by lc, say you have only a bunch of 0 in your int array
if(s_num[0].charAt(0) == ‘0‘)
return "0";
StringBuilder sb = new StringBuilder();
for(String s: s_num)
sb.append(s);
return sb.toString();
}
}
奇技淫巧:
Comparator<String> comp = new Comparator<String>(){ @Override public int compare(String str1, String str2){ String s1 = str1 + str2; String s2 = str2 + str1; return s2.compareTo(s1); // reverse order here, so we can do append() later } };
标签：大小 char public 去掉 compare ges integer compareto rri
原文地址：http://www.cnblogs.com/apanda009/p/7639483.html
