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.

题意：当输入一个字符时，这个字符可能出现多次，给出原串和显示的串，问是否相匹配

思路：维护双指针i，j，从j开始在type中找name[i],找到之后另i++，j等于找到的位置+1，然后继续找，找不到返回false。

C代码：

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代码：

扫描二维码关注公众号，回复： 6185910 查看本文章

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;
    }
}