返回零个数的递归方法

Question

这是我用来研究的一个旧测试。 我需要编写一个递归方法，返回 int[] 上从位置 0 开始的零个数。 给定 int numberOfZeroes(int[] a, int right);

Answer 1

这假设 right < a.长度

int numberOfZeroes(int[] a, int right) {
  if(right < 0) {  // We've gone through all indices
    return 0;  // So we don't want to recurse anymore
  } else if(a[right] == 0) {  // The current index has a zero
    return 1 + numberOfZeroes(a, right - 1); // Call the function, moving left one. Add one to the returned count since we found a zero
  } else {  // The current index does not have a zero
    return numberOfZeroes(a, right - 1); // Call the function, moving left one. We don't add anything since we didn't find a zero
  }
}

Answer 2

int numberOfZeroes(int[] a, int right) {
     if (right == 0) return 0;
     return numberOfZeros(a, right-1) + a[right] == 0 ? 0 : 1;
} 

执行 numberOfZeros(a, a.length) 来获取整个数组中零的数量。