Question

import QtQuick 2.0

Item {
    id: main
    visible: true
    width: 600
    height: 600
    property bool virticalFlick: false  //To get either vertical or horizontal flicking

    Flickable {
        anchors.fill: parent
        contentWidth: contentItem.childrenRect.width
        contentHeight: contentItem.childrenRect.height
        flickableDirection: Flickable.VerticalFlick
        interactive: (virticalFlick === true)?true:false
        Column {
            id: column
            spacing: 10
            Repeater {
                id: repeater
                model: 20
                ListView {
                    id: horizontalList
                    width: 600;
                    height: 100;
                    clip: true
                    interactive: (virticalFlick === true)?false:true
                    spacing: 10
                    orientation: ListView.Horizontal
                    model: 20
                    property var verticalIndex : index
                    onMovingChanged: {
                        if(moving == true) {
                            for(var i=0; i<repeater.count ; i++) {
                                /* If the property is later assigned a static value from a JavaScript statement,
                            this will remove the binding.
                            However if the intention is to create a new binding then the property
                            must be assigned a Qt.binding() value instead. This is done by passing a function to
                            Qt.binding() that returns the desired result */
                                if (i !== index)
                                    repeater.itemAt(i).contentX = Qt.binding(function() { return contentX });
                            }
                        }
                        else {
                            for(var i=0; i<repeater.count ; i++) {
                                if (i !== index)
                                    repeater.itemAt(i).contentX = contentX; // This will remove binding
                            }
                        }

                    }

                    delegate: Rectangle {
                        property var colors : ['red', 'green', 'blue']
                        property var widths : [100, 200, 300]
                        height: 100
                        width: widths[(ListView.view.verticalIndex + model.index) % widths.length]
                        color: colors[model.index % colors.length]

                        MouseArea {
                            anchors.fill: parent
                            onClicked: {
                                console.log("Rectangle.onClicked")
                            }

                        }
                    }
                }
            }
        }
    }
}

此函数查找数字的位数之和，直到它小于10。例如，如果n = 12345 然后输出= 6 如1 + 2 + 3 + 4 + 5 = 15，再次1 + 5 = 6。

Answer 1

我猜你试图达到的目标是这样的：

class St2 {
    static int sumOfDigits(int n) {
        if(n<=9) {
            return n;
        }
        else {  
            int d = n%10;
            return d + sumOfDigits((n-d)/10);
        }
    }

    public static void main(String[] args) {
        System.out.println(St2.sumOfDigits(345678));
    }
}

这将是数字的总和。复杂度在数字位数上是线性的，因此在输入数字的比例中是对数的。

以下功能的时间复杂度是多少？

1 个答案: