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.
给一组数，求这一组数的最大组合。
刚开始想用直接排序的方法：1、最高位较大的放在前面
2、但是就出现了54与5这种情况，那么进一步判断。
3、然后又出现了121与12的情况，就越写越复杂。也没有写对。
public class Solution {
public String largestNumber(int[] nums) {
System.out.println(compare(121,12));
StringBuffer str = new StringBuffer();
sort(nums, 0, nums.length - 1);
for (int i = nums.length - 1;i >= 0; i--){
System.out.print(nums[i]+" ");
str.append(nums[i]);
}
return str.toString();
}
private void sort(int[] nums, int left, int right){
if (left >= right){
return ;
}
int start = left;
int end = right;
int flag = nums[left];
while (left < right){
while (right > left && compare(flag, nums[right]) == -1){
right--;
}
if (left == right){
break;
} else {
nums[left] = nums[right];
nums[right] = flag;
left++;
}
while (right > left && compare(nums[left],flag) == -1){
left++;
}
if (left == right){
break;
} else {
nums[right] = nums[left];
nums[left] = flag;
right--;
}
}
for( int i = 0;i < nums.length; i++)
System.out.print(nums[i]+" ");
System.out.println();
sort(nums, start, left - 1);
sort(nums, right + 1, end);
}
private int compare(int num1, int num2){
double num1_copy = num1;
double num2_copy = num2;
while (num1_copy >= 10){
num1_copy = num1_copy / 10;
}
while (num2_copy >= 10){
num2_copy = num2_copy / 10;
}
if ((int) num1_copy % 10 > (int) num2_copy % 10){
return 1;
} else if ((int) num1_copy % 10 < (int) num2_copy % 10){
return -1;
} else {
int flag = (int) num1_copy % 10;
while ((int) num1_copy % 10 == (int) num2_copy % 10 && (int) num1_copy != 0 && (int) num2_copy != 0){
flag = (int) num1_copy % 10;
num1_copy = num1_copy * 10 - ((int) num1_copy % 10) * 10;
num2_copy = num2_copy * 10 - ((int) num2_copy % 10) * 10;
}
System.out.println(num1+" "+num2+" "+num1_copy+" "+num2_copy);
if ((int) num1_copy == 0 ){
if (num2_copy % 10 > flag){
return -1;
}else {
return 1;
}
} else if (num2_copy == (double) 0){
if (num1_copy % 10 > flag){
return 1;
} else {
return -1;
}
}else if (num1_copy % 10 > num2_copy % 10){
return 1;
} else {
return -1;
}
}
}
}
2、用另一种方法排序：
直接用String存储数字，两个数字(str1,str2)的大小：
s1 = str1 + str2;
s2 = str2 + str2;
s1.compareTo(s2);
这样比较。
public class Solution {
public String largestNumber(int[] nums) {
String[] strs = new String[nums.length];
for (int i = 0; i < nums.length; i++){
strs[i] = String.valueOf(nums[i]);
}
Comparator<String> comp = new Comparator<String>(){
public int compare(String str1, String str2){
String s1 = str1 + str2;
String s2 = str2 + str1;
return s2.compareTo(s1);
}
};
Arrays.sort(strs, comp);
if (strs[0].charAt(0) == '0'){
return "0";
}
StringBuffer sb = new StringBuffer();
for (String str : strs){
sb.append(str);
}
return sb.toString();
}
}