用Java8Streams实现迭代法到函数法的转换

Question

    while ((start + end) / 2 != 0) {
        Double mid = (start + end) / 2;
        if (Math2.pow(mid, 3) < n) {
            start = mid;
        } else if (Math.abs(n - Math2.pow(mid, 3)) > Math2.pow(e, 3)) {
            end = mid;
        } else if (Math.abs(n - Math2.pow(mid, 3)) < Math2.pow(e, 3)) {
            return mid;
        }
    }
    return 0.0;

}

Answer 1

Java Stream API从Java9开始就有方法Stream::iterate，因此表示迭代步骤/状态的类可以如下实现：

java prettyprint-override">class CubeSolver {
    static final double EPS = 1E-06;
        
    private double start, end, n, mid;

    public CubeSolver(double s, double e, double n) {
        this.start = s;
        this.end = e;
        this.n = n;
        this.mid = (start + end) / 2;
    }

    // UnaryOperator<CubeSolver> for iteration
    public CubeSolver next() {
        if (done()) {
            return this;
        }
        if (Math.pow(mid, 3) < n) {
            start = mid;
        } else if (Math.abs(n - Math.pow(mid, 3)) > EPS) {
            end = mid;
        }
        return new CubeSolver(start, end, n);
    }
    
    // define end of calculation
    public boolean done() {
        return mid == 0 || Math.abs(n - Math.pow(mid, 3)) < EPS;
    }
    
    @Override
    public String toString() {
        return "root = " + mid;
    }
}

然后基于流的解决方案如下所示：

1. 用开始、结束、n
定义初始种子
2. 使用stream::iterate和hasnext谓词创建有限的流2a)或使用旧的stream::iterate不使用hasnext，但使用stream::takewhile操作有条件地限制流-自Java 9
以来也可用
3. 使用stream::reduce获取流的最后一个元素

CubeSolver seed = new CubeSolver(1.8, 2.8, 8);
CubeSolver solution = Stream
    .iterate(seed, cs -> !cs.done(), CubeSolver::next)
    .reduce((first, last) -> last) // Optional<CubeSolver>
    .orElse(null);
    
System.out.println(solution); 

产出：

root = 2.0000002861022947

在Java11中，添加了静态谓词::NOT，因此使用TakeWhile的2A解决方案可以如下所示：

CubeSolver seed = new CubeSolver(0, 7, 125);
CubeSolver solution = Stream
    .iterate(seed, CubeSolver::next)
    .takeWhile(Predicate.not(CubeSolver::done))
    .reduce((first, last) -> last) // Optional<CubeSolver>
    .orElse(null);
    
System.out.println(solution); 

输出（对于EPS=1E-12):

root = 4.999999999999957