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awangdevawangdev
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minor upate
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lines changed

‎Java/Intersection of Two Arrays.java

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tags: Hash Table, Two Pointers, Binary Search, Sort
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3-
方法1: 用到hashset找unique && duplicate: O(m+n)
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方法2: 可以用binary search 找数字. Note:binary search一定需要array sorted: nLog(m)
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- 方法1: 用到hashset找unique && duplicate: O(m+n)
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- 方法2: 可以用binary search 找数字. Note:binary search一定需要array sorted: nLog(m)
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```
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/*
@@ -27,19 +29,19 @@ public int[] intersection(int[] nums1, int[] nums2) {
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if (nums1 == null || nums2 == null || nums1.length == 0 || nums2.length == 0) {
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return new int[0];
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}
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final Set<Integer> unionSet = new HashSet<>();
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final Set<Integer> resultSet = new HashSet<>();
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for (final int num: nums1) {
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Set<Integer> unionSet = new HashSet<>();
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Set<Integer> resultSet = new HashSet<>();
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for (int num: nums1) {
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unionSet.add(num);
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}
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for (final int num: nums2) {
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for (int num: nums2) {
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if (unionSet.contains(num)) {
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resultSet.add(num);
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}
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}
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int i = 0;
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final int[] result = new int[resultSet.size()];
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for (final int num: resultSet) {
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int[] result = new int[resultSet.size()];
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for (int num: resultSet) {
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result[i++] = num;
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}
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return result;

‎Java/Roman to Integer.java

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tags: Math, String
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熟悉罗马字母规则
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1. 'I V X L C D M' 分别代表的数字

‎Java/Strobogrammatic Number.java

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tags: Hash Table, Math
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根据题意枚举, 再根据规则basic implementation
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#### Alter input
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#### HashTable + Two Pointer
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```
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/*
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A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

‎Java/Valid Parentheses.java

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tags: String, Stack
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剥皮过程。解铃还须系铃人
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左边的外皮'{['在stack底部
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右边的外皮应该和stack顶上的左外皮一一对应
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7-
8-
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```
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/*
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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
@@ -19,17 +19,12 @@
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//lock will be unlocked by the same key
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//put in stack. when '),],}' appears, check stack.top() to make sure they are good match
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/*
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Use stack to hold the parentheses (head/tail) and remove them accordingly.
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*/
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// Use stack to hold the parentheses (head/tail) and remove them accordingly.
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class Solution {
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public boolean isValid(String s) {
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if (s == null || s.length() == 0) {
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return true;
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}
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final char[] charArray = s.toCharArray();
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final Stack<Character> stack = new Stack<Character>();
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for (char c : charArray) {
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if (s == null || s.length() == 0) return true;
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final Stack<Character> stack = new Stack<>();
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for (char c : s.toCharArray()) {
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if (c == '(' || c == '{' || c == '[') {
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stack.add(c);
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} else if (c == ')' || c == '}' || c == ']') {
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}
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}
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/*
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Thoughts:
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Did this on Leetcode. Think about how do we naturally check it?

‎Java/Zigzag Iterator.java

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Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?
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Clarification for the follow up question - Update (2015-09-18):
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The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:
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The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases.
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If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic".
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For example, given the following input:
2123
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[1,2,3]
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[4,5,6,7]

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