LeetCode 1055. Shortest Way to Form String

原题链接在这里：https://leetcode.com/problems/shortest-way-to-form-string/

题目：

From any string, we can form a subsequence of that string by deleting some number of characters (possibly no deletions).

Given two strings source and target, return the minimum number of subsequences of source such that their concatenation equals target. If the task is impossible, return -1.

Example 1:

Input: source = "abc", target = "abcbc"
Output: 2
Explanation: The target "abcbc" can be formed by "abc" and "bc", which are subsequences of source "abc".

Example 2:

Input: source = "abc", target = "acdbc"
Output: -1
Explanation: The target string cannot be constructed from the subsequences of source string due to the character "d" in target string.

Example 3:

Input: source = "xyz", target = "xzyxz"
Output: 3
Explanation: The target string can be constructed as follows "xz" + "y" + "xz".

Constraints:

• Both the source and target strings consist of only lowercase English letters from "a"-"z".
• The lengths of source and target string are between 1 and 1000.

题解：

Iterate through the source to move pointer in target as much as possible.

After one iteration, res++.

If we could come to the end of target, return res.

If go through one iteration, the pointer in target is not changed, then return -1.

Time Compleixty: O(n + res * m). n = target.length(). m = source.length().

Space: O(1).

AC Java:

class Solution {
    public int shortestWay(String source, String target) {
        if(source == null || target == null){
            return -1;
        }
        
        if(target.length() == 0){
            return 0;
        }
        
        int res = 0;
        int m = source.length();
        int n = target.length();
        int cur = 0;
        while(cur < n){
            int temp = cur;
            for(int i = 0; i < m && cur < n; i++){
                if(source.charAt(i) == target.charAt(cur)){
                    cur++;
                }
            }
            
            if(temp == cur){
                return -1;
            }
            
            res++;
        }
        
        return res;
    }
}

类似Is Subsequence.