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  • leetcode 58 Length of Last Word ----- java

    Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

    If the last word does not exist, return 0.

    Note: A word is defined as a character sequence consists of non-space characters only.

    For example, 
    Given s = "Hello World",
    return 5.

    这道题就是给定一个字符串,然后判断最后一个单词的长度。似乎是可以直接调用split()的。。。当然,并不是最优的。但却是最简单的,不过了解一下split()也好

    public class Solution {
        public int lengthOfLastWord(String s) {
            String[] tt;
            tt = s.split(" ");
            return tt.length == 0?0:tt[tt.length-1].length();
        }
    }

    换用正经的做法。也很简单,一次循环搞定。然而是我太天真了,从前向后必然不是最优。。。。

    public class Solution {
        public int lengthOfLastWord(String s) {
            int result = 0,len = s.length();
            for( int i = 0;i<len;i++){
                if( s.charAt(i) == ' '){
                    while( i<s.length() && s.charAt(i) == ' ')
                        i++;
                    if( i!= len)
                        result = 1;
                }
                else
                    result++;
            }        
            return result;
        }
    }

    好吧,从后往前,这样就ok了。

    public class Solution {
        public int lengthOfLastWord(String s) {
            int result = 0,i = s.length()-1;
            while( i>=0 && s.charAt(i) == ' ' ){
                i--;
            }
            while( i>=0 && s.charAt(i) != ' '){
                result++;
                i--;
            }
            return result;
        }
    }
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  • 原文地址:https://www.cnblogs.com/xiaoba1203/p/5808782.html
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