20190130

code/lc118.java

 package code;
-
+/*
+ * 118. Pascal's Triangle
+ * 题意：上层元素相邻的和，两边补1，得到下层元素
+ * 难度：Easy
+ * 分类：Array
+ * 思路：迭代计算
+ * Tips：
+ */
+import java.util.ArrayList;
 import java.util.List;
 
 public class lc118 {
     public List<List<Integer>> generate(int numRows) {
+        List<List<Integer>> res = new ArrayList<>();
+        if(numRows==0) return res;
+        List<Integer> ls = new ArrayList();
+        ls.add(1);  //第一层1个1
+        res.add(ls);
+        numRows--;
+        if(numRows==0) return res;
         while(numRows>0){
-            
+            ArrayList<Integer> temp = new ArrayList();
+            temp.add(1);
+            for (int i = 0; i < ls.size()-1 ; i++) {
+                temp.add(ls.get(i)+ls.get(i+1));
+            }
+            temp.add(1);
+            res.add(temp);
+            ls = temp;
+            numRows--;
         }
+        return res;
     }
 }

code/lc125.java

+package code;
+/*
+ * 125. Valid Palindrome
+ * 题意：判断是否为回文
+ * 难度：Easy
+ * 分类：Two Pointers, String
+ * 思路：两个指针。另一种是正则表达式替换数字，字母为空格，再反转，判断是否相等。
+ * Tips：记下另一种方法。第一种方法Bingo!
+ *       lc234, lc5
+ */
+public class lc125 {
+    public static void main(String[] args) {
+        System.out.println(isPalindrome("A man, a plan, a canal: Panama"));
+    }
+    public static boolean isPalindrome(String s) {
+        int begin = 0;
+        int end = s.length()-1;
+        s = s.toLowerCase();    //转为小写
+        while(begin<end && !(helper(s.charAt(begin)))) begin++;
+        while(begin<end && !(helper(s.charAt(end)))) end--;
+        while(begin<end){
+            if(s.charAt(begin)==s.charAt(end)){
+                begin++;
+                end--;
+                while(begin<end && !(helper(s.charAt(begin)))) begin++;
+                while(begin<end && !(helper(s.charAt(end)))) end--;
+            }else{
+                return false;
+            }
+        }
+        return true;
+    }
+
+    public static boolean helper(Character ch){ //过滤非字符数字
+        if(Character.isAlphabetic(ch)||Character.isDigit(ch)) return true;
+        return false;
+    }
+
+    public boolean isPalindrome2(String s) {
+        String actual = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
+        String rev = new StringBuffer(actual).reverse().toString();
+        return actual.equals(rev);
+    }
+}

code/lc5.java

@@ -5,6 +5,7 @@ package code;
  * 难度：Medium
  * 分类：String, Dynamic Programming
  * Tips：从后往前遍历，保证后续dp时，子情况已计算出
+ *       还有一种思路是从中间往两边扩展，中间有两种情况，一种一个字符，一种两个字符
  */
 public class lc5 {
     public static void main(String[] args) {

readme.md

@@ -75,13 +75,17 @@ Language: Java
 | 098 [Java](./code/lc98.java)
 | 101 [Java](./code/lc101.java)
 | 102 [Java](./code/lc102.java)
+| 103 [Java](./code/lc103.java)
 | 104 [Java](./code/lc104.java)
 | 105 [Java](./code/lc105.java)
 | 108 [Java](./code/lc108.java)
 | 114 [Java](./code/lc114.java)
+| 114 [Java](./code/lc116.java)
+| 114 [Java](./code/lc118.java)
 | 121 [Java](./code/lc121.java)
 | 122 [Java](./code/lc122.java)
 | 124 [Java](./code/lc124.java)
+| 125 [Java](./code/lc125.java)
 | 128 [Java](./code/lc128.java)
 | 136 [Java](./code/lc136.java)
 | 139 [Java](./code/lc139.java)