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leetcode -- 58. Length of Last Word【遍歷陣列的次序:前後】

題目

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.

題意

給定一個字串 ,返回這個字串最後一個單詞的長度。

分析及解答

解答1:(從後往前)【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:(從前往後)

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;
    }