【leetcode】solutioninjava——Easy1

public int hammingDistance(int x, int y) {int res&＃61;0,xor&＃61;x^y; for (int i &＃61; 0; i <32; i&＃43;&＃43;) {res&＃43;&＃61;(xor&1);xor&＃61;(xor>>1);} return res; }

public int findComplement(int num) {int index&＃61;0;int res&＃61;0;int tempnum&＃61;num;int curr;while(tempnum>0){curr&＃61;(tempnum&1)^1;res&＃43;&＃61;(curr<<index);tempnum&＃61;tempnum>>1;&＃43;&＃43;index;}return res;}

public String[] findWords(String[] words) {String[] lines&＃61;{"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"}; List res&＃61;new ArrayList(); for(String word:words){char[] chars&＃61;word.toUpperCase().toCharArray(); int lineindex&＃61;-1;boolean flag&＃61;true;for(int i&＃61;0;i<3;&＃43;&＃43;i){if(lines[i].indexOf(chars[0])>&＃61;0){lineindex&＃61;i; break;}}for(int i&＃61;1;ii){if(lines[lineindex].indexOf(chars[i])<0){flag&＃61;false;break;} }if(flag){res.add(word);}}
//这里注意List转换为String数组是通过 list.toArray(new String[0])实现的 return res.toArray(new String[0]); }

public class Solution { public String[] findWords(String[] words) { String[] strs &＃61; {"QWERTYUIOP","ASDFGHJKL","ZXCVBNM"}; Map map &＃61; new HashMap<>(); for(int i &＃61; 0; i){ for(char c: strs[i].toCharArray()){ map.put(c, i);} } List res &＃61; new LinkedList<>(); for(String w: words){ if(w.equals("")) continue; int index &＃61; map.get(w.toUpperCase().charAt(0)); for(char c: w.toUpperCase().toCharArray()){ if(map.get(c)!&＃61;index){ index &＃61; -1;break; } } if(index!&＃61;-1) res.add(w);} return res.toArray(new String[0]); }
}

public int[] nextGreaterElement(int[] findNums, int[] nums) {List res&＃61;new ArrayList(); for(int j:findNums){int greater&＃61;-1;int jindex&＃61;nums.length;for(int i&＃61;0;ii){if(nums[i]&＃61;&＃61;j){jindex&＃61;i;}if((nums[i]>j)&&(i>jindex)){greater&＃61;nums[i];break;}}res.add(greater);}int[] result&＃61;new int[res.size()];for(int i&＃61;0;ii){result[i]&＃61;(Integer) res.get(i);}return result;}

public int[] nextGreaterElement(int[] findNums, int[] nums) {int[] result&＃61;new int[findNums.length];//首先&＃xff0c;用一个map把nums的数值与下标记录下来Map map&＃61;new HashMap();for(int i&＃61;0;ii){map.put(nums[i], i);} for(int i&＃61;0;ii){//在遍历findnums的元素时&＃xff0c;直接通过map.get(key)找到nums中相应元素的下标作为遍历nums的起始位置int startindex&＃61;map.get(findNums[i]);for(int j&＃61;startindex;jj){if(nums[j]>findNums[i]){result[i]&＃61;nums[j];break;} result[i]&＃61;-1;} }return result;}

public int[] nextGreaterElement(int[] findNums, int[] nums) {int[] result&＃61;new int[findNums.length];Stack stack&＃61;new Stack();Map map&＃61;new HashMap();//逐个把nums入栈&＃xff0c;并且从第二个起&＃xff0c;逐个与栈顶元素比较&＃xff0c;大于它则说明此数为第一个大于栈顶元素的值&＃xff0c;记录到map去&＃xff0c;并把栈顶弹出。否&＃xff0c;则把这个元素入栈。遍历下一个for(int num:nums){//对当前nums元素&＃xff0c;用一个while处理栈中待处理的元素们&＃xff1a;栈顶<当前元素&＃xff0c;则记录到map并弹出。此时若栈未空&＃xff0c;且栈顶继续<当前nums元素&＃xff0c;则继续记录到mapwhile(!stack.empty() && stack.peek()<num){map.put(stack.pop(), num);}//把当前元素入栈&＃xff0c;在之后的循环中寻找大于它的元素
stack.push(num);} for(int i&＃61;0;ii){result[i] &＃61; map.get(findNums[i])&＃61;&＃61;null?-1:map.get(findNums[i]);}return result;}

public List fizzBuzz(int n) {List resList&＃61;new ArrayList();//求出n之内3、5的最大倍数int times_of_three&＃61;(int) Math.floor(n/3);int times_of_five&＃61;(int) Math.floor(n/5);//把1~n的3、5倍数值放到map里Map map3&＃61;new HashMap();Map map5&＃61;new HashMap(); for(int i&＃61;1;i<&＃61;times_of_three;&＃43;&＃43;i){map3.put(3*i, i);}for(int i&＃61;1;i<&＃61;times_of_five;&＃43;&＃43;i){map5.put(5*i, i);}//遍历1~n。对每个i分4种情况处理for(int i&＃61;1;i<&＃61;n;&＃43;&＃43;i){if(map3.get(i)!&＃61;null && map5.get(i)!&＃61;null){resList.add("FizzBuzz");}else if (map3.get(i)!&＃61;null && map5.get(i)&＃61;&＃61;null) {resList.add("Fizz");}else if (map3.get(i)&＃61;&＃61;null && map5.get(i)!&＃61;null) {resList.add("Buzz");}else{resList.add(""&＃43;i);} } return resList; }