摘要： class Solution { public int calculate(String s) { s = s.replace(" ", "")+'+'; //Add a '+' or '-' is necessary, otherwise, the case (1+1) will return 0 阅读全文

摘要： 这道题跟227题，背下来！ Iterative: class Solution { public int calculate(String s) { s=s.replace(" ", "")+'+'; Stack<Integer> stk = new Stack<>(); int sign =1; 阅读全文

摘要： Using Binary Search Template: class Solution { public int minEatingSpeed(int[] piles, int h) { int l=1, r=0; for(int i=0;i<piles.length;i++){ r = Math 阅读全文

摘要： class Solution { public int minFlipsMonoIncr(String s) { int oneNum = 0, flipNum =0; int res =0; for(char c: s.toCharArray()){ if(c=='0'){ if(oneNum== 阅读全文

摘要： By adding a 'max' and 'min' variable, we can make the solution's time complexity from O(n3) to O(n2): public long subArrayRanges(int[] nums) { long re 阅读全文

摘要： Solution 1, Using a PriorityQueue class Solution { public int maximumUnits(int[][] boxTypes, int truckSize) { PriorityQueue<int[]> queue = new Priorit 阅读全文

摘要： I user Trie to store the products. For every Trie node, it has the links to the next character and a PriorityQueue. The PriorityQueue is used to store 阅读全文

摘要： This is a problem which check whether you know how to user Arrays.sort(). Time complexity: O(n*log(n)) Solution 1: Using comparator class Solution { p 阅读全文

摘要： My first solution: 1. find lowest common ancestor of start and dest. 2. find path's length from LCA to start 3. find path from LCA to dest. 4. Combine 阅读全文

摘要： The following is my first solution, very easy to understand, but brute force: For every snapshot, we clone the whole array. Time complexity of snap(): 阅读全文