【242】Valid Anagram：
Given two strings s and t, write a function to determine if t is an anagram of s.
For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.
思路：看一个字符串是否是另一个字符串乱序形成的，这里面首先要考虑重复字母的情况，并且考虑时间空间复杂度，如果字符串很长的话。通过遍历其中一个字符串，逐个比较是否contains的方法不满足复杂度要求。所以考虑使用一个数组，索引是字母，值是这个字母出现的频率，两个字符串都指向这个数组，一个字符串增加频率，一个减小，然后for循环遍历这个数组，如果有非0值则返回FALSE
public class Solution {
public boolean isAnagram(String s, String t) {
if(s.length()!=t.length()){
return false;
}
int[] count = new int[26];
for(int i=0;i<s.length();i++){
count[s.charAt(i)-'a']++;
count[t.charAt(i)-'a']--;
}
for(int i:count){
if(i!=0){
return false;
}
}
return true;
}
}
【235】Lowest Common Ancestor of a Binary Search Tree
可以先参考这篇博客：http://blog.csdn.net/beiyetengqing/article/details/7633651
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself
思路：找最近公共祖先，二叉树一般用的都是递归，什么时候停止递归，由于这是二叉搜索树，所以当左右结点当一个比根节点小，一个比根节点大就停止，也就是对应以下代码中判断条件中的第三种情况，另外再比较都比根节点小或者都比根节点大的时候，用Math.max()或者Math.min()更优雅
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||p==null||q==null)return null;
if(Math.max(p.val,q.val)<root.val)return lowestCommonAncestor(root.left, p, q);//both lower than root
if(Math.min(p.val,q.val)>root.val)return lowestCommonAncestor(root.right, p, q);//both higher than root
return root;//the third cosition }
}
【191】Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.
思路：
求一个数中1的个数，也就是汉明距离，起初打算转换用string来处理，但是显然这样很慢，还有一种是按位处理，通过按位与1相与，然后记录次数，然后让这个数逻辑右移（算数右移和逻辑右移的区别：运算符“>>”执行算术右移，它使用最高位填充移位后左侧的空位，这样可以保证符号不变。右移的结果为：每移一位，第一个操作数被2除一次，移动的次数由第二个操作数确定。逻辑右移或叫无符号右移运算符“>>>“只对位进行操作，没有算术含义，它用0填充左侧的空位，不保证符号。）
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
String str = Integer.toBinaryString(n);//先把数转换成二进制
return (str.replace("0","").length());
}
}
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
return Integer.bitCount(n);//直接算一个整数的非0个数
}
}
public int hammingWeight(int n) {
int result = 0;
while (n != 0) {
if ((n & 1) == 1) {
result++;
}
n >>>= 1;
}
return result;
}
【169】Majority Element
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
找超过数组长度一半的高频元素
思路：用字符串来处理永远不是最优的方法呀，字符串只能用来转换运算，不能实现逻辑功能。想象一下木桶原则的逆原则。
public class Solution {
public int majorityElement(int[] nums) {
int major = nums[0];//以第一个元素当探针
int count = 1;
for(int i=1;i<nums.length;i++){//注意从1开始，即第二个元素开始
if(count==0){//如果出现了和这个探针的元素频率相同的数字，则探针元素重新赋值，并且次数为1，接下来进行下一循环，不再执行后面 的判断
count = 1;
major = nums[i];
}else if(major ==nums[i]){
count++;
}else if(major != nums[i]){
count--;
}
}
return major;
/*String numsStr = Arrays.toString(nums);
int size = nums.length;
int[] freq = new int[size];
int max = 0;
for(int i=0;i<size;i++){
freq[i] = size-numsStr.replace(numsStr.charAt(i)+"","").length();
if(max<freq[i]) max = freq[i];
}
return max;*/
}
}