LeetCode 161 One Edit Distance

Problem:

Given two strings S and T, determine if they are both one edit distance apart.

Analyse:

delete, add, replace 都是one edit distance.
delete 和 add: 两个string 的长度相差一
replace: 两个string的长度相同

伪代码:

string s, t
n <- s.length
m <- t.length

if n - m >= 2 || m-n >= 2
return false
if n-m = 1 || m-n =1
return isOneChangeDitance()
if n-m=0
return isOneReplaceDistance()

下面讨论一些细节
isOneChangeDitance()

s: z x c v b
l
t: z x v b
s

对于这个列子,可以看出delete s 的一个char或者给 t add一个char 就相同. 所以是One Edit Distance.
In this problem, we can create two pointer. One is long string’s pointer: l. Another one is short string’s pointer: s. Both of two pointers go with the same pace to scan the string. When they point to the same character, both of them move to the next character. When they point to the different character, the long pointer move forward and the short one stay. After that, they go with the same pace to check what’s going on in the following parts of the string.

isOneReplaceDistance()

s: z x c b
l
t: z x v b
s

In thsi scenario, we also make l pointer and s pointer go with the same pace. When they find the different character, count it and move forward. If the counting number is larger than 1, return false.

其他

1. 当两个string完全相同时,也不算做 One Edit Distance
2. 注意如何建立两个指针同时扫面Sting
3. 还有一些Conner Situation

code(Java)

public class Solution1 {

	public static boolean isOneEditDistance(String s, String t) {
		
		if(s == null && t == null) return true;
		if(s == null || t == null) return false;
		
		int n = s.length();
		int m = t.length();
		
		
		if (n - m >= 2 || m - n >= 2) return false;
		
		if (n - m == 1) {
			return isOneChangeDistance(s,t);
		}
		
		if ( m - n == 1) {
			return isOneChangeDistance(t,s);
		}
		
		if ( n - m == 0) {
			return isOneReplaceDistance(s,t);
		}
		
		return false;
	}
	
//	default string s is longer than string t
	public static boolean isOneChangeDistance(String s, String t) {
		
		int count = 0;
		//扫描长String比较安全,扫描短的容易out of bound
		for (int i=0, j=0; i<s.length(); i++) {
			if (s.charAt(i) == t.charAt(j)) {
				j++;
			}else {
				count++;
			}
		}	
		if (count == 1) return true;
		else return false;

	}
	
	
	public static boolean isOneReplaceDistance(String s, String t) {
		
		int count = 0;
		for (int i=0; i<t.length(); i++) {
			if (s.charAt(i) != t.charAt(i)) {
				count++;
			}		
		}
		if (count == 1) return true;
		else return false;
	}
	
	
	
	public static void main(String[] args) {
		
		String s = "";
		String t = "";		
		System.out.println(isOneEditDistance(s,t));
		
	}		
}