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其实就是把有道云笔记的东西搬过来，
不定期会更新
随便写写的东西，很多都需要优化，
如果有人看到，希望能给点建议，谢谢！
public class Solution {
public boolean isPalindrome(int x) {
int y = 0;
int tmp = x;
if(x < 0){tmp = x*(-1);}
while(tmp != 0){
y = y*10 + tmp%10;
tmp = tmp/10;
}
if(x == y){
return true;
}
return false;
}
}
Given a string s consists of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
For example,
Given s = "Hello World",
return 5.
思路是从后往前，如果有字符，就++，如果是空，判断count是否有值
public class Solution {
public int lengthOfLastWord(String s) {
int count = 0;
char[] wholeSentence = s.toCharArray();
for(int i = wholeSentence.length - 1; i >= 0 ; i--){
if(wholeSentence[i] != ‘ ‘){
count ++;
}
else{
if(count != 0){
return count;
}
}
}
return count;
}
}
3.power of two
public class Solution {
public int lengthOfLastWord(String s) {
int count = 0;
char[] wholeSentence = s.toCharArray();
for(int i = wholeSentence.length - 1; i >= 0 ; i--){
if(wholeSentence[i] != ‘ ‘){
count ++;
}
else{
if(count != 0){
return count;
}
}
}
return count;
}
}
to judge whether a integer is power of two
an integer can divide two totally util it becomes 1 if it is power of two
4.Summary Ranges 【需要重新考虑的问题】
Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
1 public class Solution { 2 3 public List<String> summaryRanges(int[] nums) { 4 5 List<String> result = new ArrayList<String>(); 6 7 List<Integer> temp = new ArrayList<Integer>(); 8 9 if(nums.length > 0){ 10 11 for(int i = 0; i< nums.length; i++){ 12 13 temp.add(nums[i]); 14 15 if(i != (nums.length -1) && nums[i + 1] == nums[i] + 1){ 16 17 continue; 18 19 } 20 21 else{ 22 23 result.add(generateStr(temp)); 24 25 temp.clear(); 26 27 } 28 29 } 30 31 } 32 33 return result; 34 35 } 36 37 38 39 private String generateStr(List<Integer> temp){ 40 41 if(temp.size() != 1){ 42 43 String generatedStr = temp.get(0).toString() + "->" + temp.get(temp.size()-1).toString(); 44 45 return generatedStr;} 46 47 else{ 48 49 return temp.get(0).toString(); 50 51 } 52 53 } 54 55 }
取一个区间，放在一个list，然后把这个list转换成一个string，这个应该算是比较笨的办法，仔细想想怎么做会更加好？
5.valid anagram
1 Given two strings s and t, write a function to determine if t is an anagram of s. 2 3 For example, 4 s = "anagram", t = "nagaram", return true. 5 s = "rat", t = "car", return false. 6 7 Note: 8 You may assume the string contains only lowercase alphabets. 9 10 public class Solution { 11 12 public boolean isAnagram(String s, String t) { 13 14 15 16 Map<Character, Integer> sHash = new HashMap<Character, Integer>(); 17 18 Map<Character, Integer> tHash = new HashMap<Character, Integer>(); 19 20 21 22 char[] sArray = s.toCharArray(); 23 24 char[] tArray = t.toCharArray(); 25 26 27 28 if(sArray.length != tArray.length) 29 30 return false; 31 32 for(int i = 0; i < sArray.length; i++){ 33 34 if(sHash.containsKey(sArray[i])){ 35 36 int count = sHash.get(sArray[i]); 37 38 sHash.put(sArray[i], ++count); 39 40 } 41 42 else{ 43 44 sHash.put(sArray[i], 1); 45 46 } 47 48 if(tHash.containsKey(tArray[i])){ 49 50 int count = tHash.get(tArray[i]); 51 52 tHash.put(tArray[i], ++count); 53 54 } 55 56 else{ 57 58 tHash.put(tArray[i], 1); 59 60 } 61 62 } 63 64 65 66 return sHash.equals(tHash); 67 68 } 69 70 71 72 }
这里使用了hashmap的比较方法，应该来讲，效率比较高
6.
contains duplicate 2
Given an array of integers and an integer k, find out whether there there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between iand j is at most k.
1 public class Solution { 2 3 public boolean containsNearbyDuplicate(int[] nums, int k) { 4 5 if(nums.length == 0 ) return false; 6 7 if(k == 0) return false; 8 9 Map<Integer,List<Integer>> intHash = new HashMap<Integer, List<Integer>>(); 10 11 for(int i = 0; i< nums.length; i++){ 12 13 if(intHash.containsKey(nums[i])){ 14 15 List<Integer> indexList = intHash.get(nums[i]); 16 17 indexList.add(i); 18 19 intHash.put(nums[i], indexList); 20 21 } 22 23 else{ 24 25 List<Integer> indexList = new ArrayList<Integer>(); 26 27 indexList.add(i); 28 29 intHash.put(nums[i], indexList); 30 31 } 32 33 } 34 35 36 37 if(!intHash.isEmpty()){ 38 39 for(Integer i: intHash.keySet()){ 40 41 List<Integer> indexList = intHash.get(i); 42 43 for(int j = 0; j< indexList.size() - 1; j++){ 44 45 for(int m = j+1; m < indexList.size(); m++){ 46 47 if((indexList.get(m) - indexList.get(j)) <= k && (indexList.get(m) - indexList.get(j)) >=(0-k) ){ 48 49 return true; 50 51 } 52 53 } 54 55 } 56 57 } 58 59 } 60 61 62 63 return false; 64 65 } 66 67 }
第一个版本的算法是：从头开始取数，然后在这个数的k区间内取别的数，进行对比，判断是否有一样的，一样就返回true，没有的话最后返回false
但是在处理[1-29999]，k值为29999的时候出现了问题，是因为没有重复的数字，全部遍历了一遍，时间复杂度为o(n*n)，造成OJ上面的runtime超时。
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原文地址：http://www.cnblogs.com/michael-faust/p/4701189.html
