查找最右边的字符串

Question

我正在编写一个递归算法，以在字符列表中查找单词“yes”的最右边出现。

public class Question7 {


    public static int where(char[] A, String s, int i) {
        // A recursive function where()
        // Return the location of rightmost occurence of a given string s in a given array A
        // Complete the where function
        // You may want to add more parameters to the where function
        s = "yes";
        where(A, s, A.length);
        if (A.length < 3) {
            return -1;
        } else if (A.length == 3) {
            if (A[i - 2] == s.charAt(0) && A[i - 1] == s.charAt(1) && A[i] == s.charAt(2)) {
                return i - 2;
            } else {
                return -1;
            }
        } else {
            if (A[i - 2] == s.charAt(0) && A[i - 1] == s.charAt(1) && A[i] == s.charAt(2)) {
                return i - 2;
            } else {
                return where(A, s, i - 1);
            }
        }

    }

    public static void main(String[] args) {
        char[] givenarray = {'o', 't', 'z', 'y', 'e', 's', 'v', 'g', 'r', 'a', 'y', 'e', 's'};

        // Test your method
        System.out.println("The rightmost occurence of 'yes' in the given array is at index " + where());
        // Your method should return 10
   }

}

调用该方法时，我的问题在底部。我应该使用特定的参数还是不特定的参数？例如： where(givenarray, "yes", givenarray.length) 或只是 (char[] A, String s, int i)？我从来都不擅长调用带参数的方法，因此不胜感激！

Answer 1

首先，您需要了解要返回的内容。

-- 所以这里你必须得到一个单词最后出现的索引。该怎么办？让我们捕捉它。有很多方法可以做到这一点。一种方法是 -

    public class Question7 {
        public static int lastFoundIndex = -1;
        ..
    }

二、如何启动流程？

    public static void main(String... args) {
        char[] givenArray = {'o', 't', 'z', 'y', 'e', 's', 'v', 'g', 'r', 'a', 'y', 'e', 's'};

        // this is the initializing function, where you pass 2 params
        // The array and Index from where you want to start looking.  
        where(givenArray, 0);  
        System.out.println("The rightmost occurence of 'yes' in the given array is at index "+ lastFoundIndex);
    }

接下来，让我们看看where(..) 函数会是什么样子。

    public static void where(char[] arr, int startIndex) {
        // this is the base case. First statement. 
        // Basically for testing, if the array is empty or we reached at the end of the execution (we'll reach there soon)
        if(startIndex >= arr.length) {
            return;
        }

        // Now we check if we have the word 'yes', if so - save the lastFoundIndex. and call where function and tell it to look starting from currentIndex + 3 place.  
        if(arr[startIndex] == 'y' && startIndex + 2 < arr.length) {
            if(arr[startIndex + 1] == 'e' && arr[startIndex + 2] == 's') {
                lastFoundIndex = startIndex;
                where(arr, startIndex+3);
            }
        }

        // if we dont find the character y, then just go on with next char.  
        where(arr, startIndex+1);
    }