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 is9534330
.
Note: The result may be very large, so you need to return a string instead of an integer.
[分析]
按题目的要求, 重写排序方法, 然后排序即可.
[注意]
当一个数是另一个数的前缀时, 要注意顺序. 这里用了个小技巧 比较 S1+S2 ? S2+S1 即可.
另外, leading zeros要去掉.
[CODE]
second try:
public class Solution { public String largestNumber(int[] arr) { Integer[] array = new Integer[arr.length]; int i = 0; for (int value : arr) { array[i++] = Integer.valueOf(value); } Arrays.sort(array, new Comparator<Integer>() { @Override public int compare(Integer a1, Integer a2) { int l1 = a1==0? 1 : (int) Math.log10(Math.abs(a1)) + 1; int l2 = a2==0? 1 : (int) Math.log10(Math.abs(a2)) + 1; long aa1 = (long) (a1 * Math.pow(10, l2) + a2); long aa2 = (long) (a2 * Math.pow(10, l1) + a1); return aa1>aa2 ? -1 : (aa1==aa2 ? 0 : 1); } } ); StringBuilder sb = new StringBuilder(); for(Integer e : array) { sb.append(e); } return sb.toString().replaceFirst("^0+(?!$)", ""); } }
first edition:
public class Solution { public String largestNumber(int[] arr) { String[] strs = new String[arr.length]; for(int i=0; i<arr.length; i++) { strs[i] = Integer.toString(arr[i]); } Arrays.sort(strs, new Comparator<String>() { @Override public int compare(String s1, String s2) { String ss1 = s1 + s2; String ss2 = s2 + s1; int i=0; while(i<ss1.length()) { if(ss1.charAt(i)!=ss2.charAt(i)) { return ss1.charAt(i) - ss2.charAt(i); } ++i; } return 0; } } ); StringBuilder sb = new StringBuilder(); for(int i=strs.length-1; i>=0; i--) { sb.append(strs[i]); } return sb.toString().replaceFirst("^0+(?!$)", ""); } }