LeetCode 440. K-th Smallest in Lexicographical Order (Java版; Hard) 题目描述 Given integers n and k, find the lexicographically k-th smallest integer in the range from 1 to n. Note: 1 ≤ k ≤ n ≤ 109. Example: Input: n: 13 k: 2 Output: 10 Explanation: The lexicographical order ...
newComparator<Integer>() {10@Override11publicintcompare(Integer i1, Integer i2) {12returnString.valueOf(i1).compareTo(String.valueOf(i2));13}14});15returnlist;16}17}
* Java 8 users: Use {@link Comparators#lexicographical(Comparator)} instead. * * @since 2.0 */ @GwtCompatible(serializable = true) // type parameter <S> lets us avoid the extra <String> in statements like: // Ordering<Iterable<String>> o = // Ordering.<String>natural().lexicographical...
Example: Program to Sort Strings in Dictionary Order fun main(args: Array<String>) { val words = arrayOf("Ruby", "C", "Python", "Java") for (i in 0..2) { for (j in i + 1..3) { if (words[i].compareTo(words[j]) > 0) { // swap words[i] with words[j[ val temp...
Java实现 1classSolution {2publicString lastSubstring(String s) {3intleft = 0;4intright = left + 1;5intstep = 0;6while(right + step <s.length()) {7if(s.charAt(left + step) < s.charAt(right +step)) {8left =right;9right++;10step = 0;11}elseif(s.charAt(left + step) ==...
方法2:来自leetcode discuss https://discuss.leetcode.com/topic/55377/simple-java-dfs-solution The idea is pretty simple. If we look at the order we can find out we just keep adding digit from0 to9 to every digit and make it a tree. Then we visit every node in pre-order. ...
本文整理了Java中com.google.common.collect.LexicographicalOrdering.<init>()方法的一些代码示例,展示了LexicographicalOrdering.<init>()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。LexicographicalOrdering.<init>()...
1163 Last Substring in Lexicographical Order 按字典序排在最后的子串 Description: Given a string s, return the last substring of s in lexicographical order. Example: Example 1: Input: s = "abab" Output: "bab" Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba",...
提示1 Assume that the answer is a sub-string from index i to j. If you add the character at index j+1 you get a better answer.提示2 The answer is always a suffix of the given string.提示3 Since the limits are high, we need an efficient data structure....