Question

所以我有这个堆栈，我不知道如何将它的内容复制到队列然后反之亦然（当从队列复制回堆栈时，值会向后倒？）

Stack st = new Stack();

st.Push('E');
st.Push('L');
st.Push('P');
st.Push('M');
st.Push('A');
st.Push('X');
st.Push('E');

Answer 1

Stack实施ICollection，因此您可以Queue ICollection Queue queue = new Queue(st);使用Stack st2 = new Stack(queue);：

start = function () {
    this.data("ox", +this.getBBox().cx);
    this.data("oy", +this.getBBox().cy);
},
move = function (dx, dy) {
    var tmpPt = {
        x: this.data('ox') + dx,
        y: this.data('oy') + dy
    };
    // move will be called with dx and dy
    l = gradSearch(l, tmpPt);
    pt = path.getPointAtLength(l);

    if (!isNaN(pt.x) && !isNaN(pt.y)) {
        this.transform('t' + (pt.x - this.data("ox")) + ',' + (pt.y - this.data("oy")));
    };
},
end = function () {
    console.log('End of drag');
}

同样适用于Stack：

import numpy as np
import scipy as sc

import matplotlib.pyplot as plt

x = np.linspace (0, 10, 100)
y = .5 * x + 4

plt.figure ()


yres = 100
ymax = np.max (y)
ymin = 0 
yy = np.linspace (ymin, ymax, yres)

fill_n = 10

xres = len(x)

# gradient image
gI = np.zeros ((yres, xres))
for xi,xx in enumerate(x):
  ym = y[xi]

  # find elment closest to curve
  ya = np.argmin (np.abs(yy - ym))

  gI[ya-fill_n:ya, xi] = np.linspace (0, 1, fill_n)

# make alpha cmap out of gray map
bb = np.linspace (0, 1, fill_n)
kk = []
for b in bb:
  kk.append ((b, b, b))

bb = tuple (kk) 
gr = { 'blue' : bb,
       'red' : bb,
       'green': bb,
       'alpha': bb }

plt.register_cmap (name = 'GrayAlpha', data = gr)

gI = np.flipud (gI)
plt.imshow (gI, vmin = 0, vmax = 1, cmap = 'GrayAlpha', interpolation = 'bicubic')
plt.show ()

如何将堆栈的内容复制到队列中并从队列中返回到c＃中的堆栈？

1 个答案: