408. Valid Word Abbreviation
作者:互联网
This is two points problem, just concentrate and carefully deal with the characters, then the problem can be solved.
This is a very good problem for writing all kinds of edge test cases.
public boolean validWordAbbreviation(String word, String abbr) { int m = word.length(), n = abbr.length(); int i = 0, j = 0; while (i < m && j < n) { if (word.charAt(i) == abbr.charAt(j)) { i++; j++; } else { if (!Character.isDigit(abbr.charAt(j)) || abbr.charAt(j) == '0') return false; else { int start = j; while (j < n && Character.isDigit(abbr.charAt(j))) { //be careful, need ot add a j < n, otherwise, eg. abbr="a1" will fail j++; } int len = Integer.valueOf(abbr.substring(start, j)); i += len; } } } return i == m && j == n; //if simply return true, eg. word = "a", abbr="2" will fail }
标签:Word,charAt,int,return,Abbreviation,Valid,abbr,&&,word 来源: https://www.cnblogs.com/feiflytech/p/15862406.html