lc q65 round 1

Author: yihaoye

Leetcode Practices/README.md

@@ -119,7 +119,7 @@
 - [x] 157	Read N Characters Given Read4		
 - [ ] 158	Read N Characters Given Read4 II - Call multiple times		
 - [x] 68	Text Justification		
-- [ ] 65	Valid Number		
+- [x] 65	Valid Number		
 ### Substring			
 - [x] 76	Minimum Window Substring		                                        Sliding Window
 - [ ] 30	Substring with Concatenation of All Words		                        Sliding Window
@@ -564,7 +564,7 @@
 - [x] 60	Permutation Sequence		
 - [x] 62	Unique Paths		
 - [x] 64	Minimum Path Sum		
-- [ ] 65	Valid Number		
+- [x] 65	Valid Number		
 - [x] 66	Plus One		
 - [x] 67	Add Binary		
 - [x] 68	Text Justification		

Leetcode Practices/algorithms/hard/65 Valid Number.java

@@ -0,0 +1,104 @@
+/*
+Given a string s, return whether s is a valid number.
+
+For example, all the following are valid numbers: "2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789", while the following are not valid numbers: "abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53".
+
+Formally, a valid number is defined using one of the following definitions:
+
+An integer number followed by an optional exponent.
+A decimal number followed by an optional exponent.
+An integer number is defined with an optional sign '-' or '+' followed by digits.
+
+A decimal number is defined with an optional sign '-' or '+' followed by one of the following definitions:
+
+Digits followed by a dot '.'.
+Digits followed by a dot '.' followed by digits.
+A dot '.' followed by digits.
+An exponent is defined with an exponent notation 'e' or 'E' followed by an integer number.
+
+The digits are defined as one or more digits.
+
+ 
+
+Example 1:
+
+Input: s = "0"
+
+Output: true
+
+Example 2:
+
+Input: s = "e"
+
+Output: false
+
+Example 3:
+
+Input: s = "."
+
+Output: false
+
+ 
+
+Constraints:
+
+1 <= s.length <= 20
+s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.
+*/
+
+
+
+// My Solution:
+class Solution {
+    public boolean isNumber(String s) {
+        // String + Divide-and-conquer
+        // recommand to forward to DFA solution
+        int ePos = -1;
+        for (int i=0; i<s.length(); i++) {
+            if (s.charAt(i) != 'e' && s.charAt(i) != 'E') continue;
+            ePos = i; break;
+        }
+        if (ePos == -1) return isDecimal(s);
+        if (s.length() == 1) return false;
+        String s1 = s.substring(0, ePos), s2 = s.substring(ePos + 1);
+        return isDecimal(s1) && isInteger(s2);
+    }
+
+    public boolean isDecimal(String s) { // include isInteger
+        if (s.isBlank()) return false;
+        int dotPos = -1;
+        for (int i=0; i<s.length(); i++) {
+            if (s.charAt(i) != '.') continue;
+            dotPos = i; break;
+        }
+        if (dotPos == -1) return isInteger(s);
+        if (s.length() == 1) return false;
+        String s1 = s.substring(0, dotPos), s2 = s.substring(dotPos + 1);
+        if ((isInteger(s1) || isSign(s1) || s1.isBlank()) && isPureInteger(s2)) return true;
+        if (isInteger(s1) && s2.isBlank()) return true;
+        return false;
+    }
+
+    public boolean isInteger(String s) {
+        if (s.isBlank()) return false;
+        for (int i=0; i<s.length(); i++) {
+            if (i == 0 && (s.charAt(i) == '-' || s.charAt(i) == '+') && s.length() != 1) continue;
+            if ('0' <= s.charAt(i) && s.charAt(i) <= '9') continue;
+            return false;
+        }
+        return true;
+    }
+
+    public boolean isPureInteger(String s) {
+        if (s.isBlank()) return false;
+        for (int i=0; i<s.length(); i++) {
+            if ('0' <= s.charAt(i) && s.charAt(i) <= '9') continue;
+            return false;
+        }
+        return true;
+    }
+
+    public boolean isSign(String s) {
+        return s.equals("+") || s.equals("-");
+    }
+}