leetcode -- 58. Length of Last Word【遍历数组的次序：前后】

题目

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.

题意

给定一个字符串 ，返回这个字符串最后一个单词的长度。

分析及解答

解答1：（从后往前）【更快】

public class Solution {    public int lengthOfLastWord(String s) {        if(s.equals("")) return 0;        char[] array = s.toCharArray();        boolean isPreSpace = true;        boolean isCurrentSpace = false;        int count = 0;        for(int i = array.length-1; i >= 0 ;i--){            isCurrentSpace = (array[i] == ' ');            if(isPreSpace){                if(!isCurrentSpace){                    count =1;                }            }else{                if(!isCurrentSpace){                    count ++;                }else{                    break;                }            }            isPreSpace = isCurrentSpace;        }        return count;    }}
解法2：（从前往后）
public int lengthOfLastWord(String s) {        if(s.equals("")) return 0;        char[] array = s.toCharArray();        boolean isPreSpace = true;        boolean isCurrentSpace = false;        int count = 0;        for(char ch : array){            isCurrentSpace = (ch == ' ');            if(isPreSpace){                if(!isCurrentSpace){                    count =1;                }            }else{                if(!isCurrentSpace){                    count ++;                }            }            isPreSpace = isCurrentSpace;        }        return count;    }