Your friend is typing his name into a keyboard.  Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.

You examine the typed characters of the keyboard.  Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.

Example 1:

Input: name = "alex", typed = "aaleex"
Output: true
Explanation: 'a' and 'e' in 'alex' were long pressed.

Example 2:

Input: name = "saeed", typed = "ssaaedd"
Output: false
Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.

Example 3:

Input: name = "leelee", typed = "lleeelee"
Output: 
 
true

Example 4:

Input: name = "laiden", typed = "laiden"
Output: true
Explanation: It's not necessary to long press any character.

Note:

1. name.length <= 1000
2. typed.length <= 1000
3. The characters of name and typed are lowercase letters.

題意：當輸入一個字元時，這個字元可能出現多次，給出原串和顯示的串，問是否相匹配

思路：維護雙指標i，j，從j開始在type中找name[i],找到之後另i++，j等於找到的位置+1，然後繼續找，找不到返回false。

C程式碼：

bool isLongPressedName(char* name, char* typed) {
    if(strcmp(name,typed) == 0) return true;
    int len1 = strlen(name);
    int len2 = strlen(typed);
    int i,j,pos;
    pos = 0;
    for(i = 0; i < len1; i++) {
        bool flag = false;
        for(j = pos; j < len2; j++) {
            if(name[i] == typed[j]) {
                pos = j + 1;
                flag = true;
                break;
            }
        }
        if(!flag) {
            break;
        }
    }
    if(i == len1) return true;
    else return false;
}

Java程式碼：

public class Solution {
	public boolean isLongPressedName(String name, String typed) {
        if(name.equals(typed)) return true;
        int pos = 0,i,j;
        int len1 = name.length();
        int len2 = typed.length();
        for(i = 0; i < len1; i++) {
        	boolean flag = false;
        	for(j = pos; j < len2; j++) {
        		if(typed.charAt(j) == name.charAt(i)) {
        			pos = j + 1;
        			flag = true;
        			break;
        		}
        	}
        	if(!flag) break;
        }
        if(i == len1) return true;
        else return false;
    }
}