leetcode 386. Lexicographical Numbers
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Given an integer n, return 1 - n in lexicographical(字典序) 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 List<Integer> lexicalOrder(int n) {List<Integer> list=new ArrayList<Integer>();for(int i=1;i<=9;i++){helper(i, n, list);}return list;}public void helper(int number,int n,List<Integer> list){if(number>n){return;}else{list.add(number);for(int i=0;i<=9;i++){helper(10*number+i, n, list);}}return;}大神则用了时间复杂度为O(n) , 空间复杂度为O(1) 的迭代方法。
public List<Integer> lexicalOrder(int n) { List<Integer> list = new ArrayList<>(n); int curr = 1; for (int i = 1; i <= n; i++) { list.add(curr); if (curr * 10 <= n) { curr *= 10; } else if (curr % 10 != 9 && curr + 1 <= n) { curr++; } else { while ((curr / 10) % 10 == 9) { curr /= 10; } curr = curr / 10 + 1; } } return list;}思路是找到:往 list 中加的后一个元素。
以当前数字为45来举个例子: 如果当前数字是45,那么下一个数字将是450 (450 == 45 * 10)(if 450 <= n), 或者 46 (46 == 45 + 1) (if 450>n and 46 <= n) 或者 5 (5 == 45 / 10 + 1) (因为5<45,所以肯定5<n).
又以 n = 600 来举个例子, 如果当前数字是 499, 那么下一个数字是5,因为在499中 "4" 后面全是 “9”,所以要一直用10来除499,直到下一个数字不是9。
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