摘要： Note: 1. Find each steps lowest cost. 2. Find whether it could reach to end or not. 阅读全文

摘要： class Solution { public int lengthLongestPath(String input) { if (input.length() == 0) return 0; String[] layers = input.split("\n"); int[] stack = new int[layers.length +... 阅读全文

摘要： 1. Use priority queue. Need to check whether one element has been double counted: 2 Binary search: For this kind of matrix, binary search should work 阅读全文

摘要： Resevior Sampling : https://en.wikipedia.org/wiki/Reservoir_sampling 阅读全文

摘要： Note: 1. Find a increasing digits number. It's kind of longest increasing subsequence but with fixed size. 2. Remember to remove the zeros from beginn 阅读全文

摘要： This is SO AWESOME solution. 阅读全文

摘要： Note: 1. 0 could be 10 but not valid for 01. Check edge cases. 2. No number shoud be there as num reduced. 阅读全文

摘要： Note: end should apply end of left to right order using i index start is right to left using length - i - 1 index. Set end to be -2 since result is en 阅读全文

摘要： Almost same as Longest Common SubString 阅读全文

摘要： Since it required exactly one modification, it is better to use following solution. Other ise, the Trie solution can work. (At least one modify). 阅读全文