简易遍历寻找0/1串中连续10个1首次出现位置(递归)-优快云博客

本文探讨了一个Java程序的执行流程，通过递归逻辑实现特定功能，详细解析了程序中字符串处理、条件判断和递归调用的过程。

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//A1.java

public class A1 {

    private final static int N = 10;

    public static void main(String[] args) {
        String str = "0010000110111101011010101000101001110111000110010001111010111100011000011111111110100100001111110001101011010001000101011100011100111000110010010111100101101011100010110100111010000001000101110101111000001111100110101010010101011011010110011010101000111011001001110111000011001110111001100010001001010100111001111111110100000110001100010101101000011111011000011111111101100000010000000111111000000101111011110101101110101101100100001011110110010110111001110110001011110000111100000000010011101100101010101100101110000011100100010000101100101001001001000101001100010100011000110001011010101010000101000111000101011100111011101101110010100110001011011100110001110010010010101101111001101101100100100011001110111111110100001001001110101101100111101101000110000101000111000001011111010000110000011110001111100100010011100001010111011110011011101010101011011101100011110001101011110100011100011100110111110010101110011101000000011100111000111001001101000011100011001110100110110100111100111011000000011000";
        System.out.println("index:" + exec(str, 0, 0));
    }

    public static int exec(String str, int i, int count) {
        if (count == N) {
            System.out.println("i:" + i);
            return i - 10;
        }
        if (i == str.length()) {
            return -1;
        }
        if (str.charAt(i) == '0') {
            return exec(str, i + 1, 0);
        } else {
            return exec(str, i + 1, count + 1);
        }
    }
}