Largest Number @leetCode

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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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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Have you met this question in a real interview?

思路：第一想法是类似于premutation II 的解法，用dfs可以解决，但肯定超时。

于是想到了利用 heap 来做，关键点在于comparator，直接比较会因为有长度不同的原因导致有错误case

后来想了好久，利用 s1 + s2 和 s2 + s1 来进行比较，可以完美的解决这个问题，因为他的本质比较就是相加后比较。

public class Solution {    public String largestNumber(int[] num) {        if (num.length < 0)return "";Comparator<String> comparator = new Comparator<String>() {public int compare(String s1, String s2) {String l1 = s1 + s2;String l2 = s2 + s1;for (int i = 0; i < l1.length(); i++) {if (l1.charAt(i) > l2.charAt(i))return -1;else if (l1.charAt(i) < l2.charAt(i))return 1;}return 0;}};Queue<String> queue = new PriorityQueue<String>(num.length, comparator);for (int i : num)queue.add(String.valueOf(i));StringBuilder sb = new StringBuilder();while (!queue.isEmpty())sb.append(queue.poll());if(sb.charAt(0) == '0'){//All 0s situation    sb = (new StringBuilder()).append('0');}return sb.toString();    }}

超时解法：

public String largestNumber(int[] num) {if (num.length < 0)return "";Arrays.sort(num);String[] ret = { "0" };permutation(num, 0, ret);return ret[0];}private void permutation(int[] num, int start, String[] ret) {if (start >= num.length) {StringBuilder sb = new StringBuilder();for (int i : num)sb.append(i);String s = sb.toString();if (ret[0].compareTo(s) < 0) {ret[0] = s;}return;}HashSet<Integer> visited = new HashSet<Integer>();for (int i = start; i < num.length; i++) {if (!visited.contains(num[i])) {swap(num, i, start);permutation(num, start + 1, ret);swap(num, i, start);visited.add(num[i]);}}}

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