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_386.java
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package com.fishercoder.solutions;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Date;
import java.util.List;
/**
* Given an integer n, return 1 - n in lexicographical order.
For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].
Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.*/
public class _386 {
//Radix sort doesn't apply here! Don't confuse myself!
//rewrote their solution from Python to Java:https://discuss.leetcode.com/topic/54986/python-memory-limit-exceeded-for-problem-386/17
public static List<Integer> lexicalOrder(int n) {
List<Integer> result = new ArrayList();
int i = 1;
while (true) {
result.add(i);
if (i * 10 <= n) {
i *= 10;
} else {
while (i % 10 == 9 || i == n) {
i /= 10;
}
if (i == 0) {
return result;
}
i++;
}
}
}
//someone on Discuss hinted that you could use recursion to solve it in Java
//then I wrote the following method, eventually, got all test cases produce the right output, but unfortunately TLE'ed by OJ
public static List<Integer> lexicalOrder_LTE_by_10458(int n) {
List<Integer> result = new ArrayList();
int insertPosition = 0;
return addNumbers(result, 1, insertPosition, n);
}
private static List<Integer> addNumbers(List<Integer> result, int insertNumber, int insertPosition, int n) {
int i;
for (i = 0; i < 9; i++) {
if (insertNumber + i > n) {
return result;
}
result.add(insertPosition + i, insertNumber + i);
if ((insertNumber + i) % 10 == 0 && (insertNumber + i) == (insertNumber + 10)) {
break;
}
}
while ((insertNumber + i) % 10 != 0 && (insertNumber + i) <= n) {
result.add(insertPosition + i, insertNumber + i);
i++;
}
//find next insert position:
insertPosition = result.indexOf((insertNumber + i) / 10);
return addNumbers(result, insertNumber + i, insertPosition + 1, n);
}
public static void main(String... strings) {
long lStartTime = new Date().getTime();
// CommonUtils.printList(lexicalOrder_TLE_by_23489(23489));
// List<Integer> result = lexicalOrder(1);//right
// List<Integer> result = lexicalOrder(13);//right
// List<Integer> result = lexicalOrder_LTE_by_10458(58);
// List<Integer> result = lexicalOrder(120);//right
// List<Integer> result = lexicalOrder(1200);
// List<Integer> result = lexicalOrder(10);
// List<Integer> result = lexicalOrder(5000000);
// List<Integer> result = lexicalOrder_LTE_by_10458(50000);//this can finish in 183 ms
List<Integer> result = lexicalOrder_LTE_by_10458(500000);
// List<Integer> result = lexicalOrder_LTE_by_10458(14959);
long lEndTime = new Date().getTime();
long difference = lEndTime - lStartTime;
System.out.println("Elapsed milliseconds: " + difference);
System.out.println("result size is: " + result.size());
// CommonUtils.printList(result);
}
/**
* The most naive way is to generate this list, sort it using a customized comparator and then return it.
* Unfortunately, this results in TLE with this input: 23489
*/
public static List<Integer> lexicalOrder_TLE_by_23489(int n) {
List<Integer> result = new ArrayList();
for (int i = 1; i <= n; i++) {
result.add(i);
}
Collections.sort(result, new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
String s1 = String.valueOf(o1);
String s2 = String.valueOf(o2);
return s1.compareTo(s2);
}
});
return result;
}
}