C# 实现颜色的梯度渐变案例

 public void drawrect(gasconcentration[] data)
    {
      graphics graphic = picturebox1.creategraphics();
      graphics graphic2 = picturebox2.creategraphics();
      int icall2 = picturebox2.width/10;
           
      data = new gasconcentration[40];
      int ilen = picturebox1.width = 540; 
      int ihigh = picturebox1.height;
      //初始化十种颜色
      color[] color = new color[10] { color.fromargb(240, 0, 0), color.green, color.yellow, color.blue, color.steelblue, color.seagreen,
                    color.chartreuse, color.saddlebrown, color.violet, color.burlywood};
          
      //十个颜色，每个颜色三个深度
      for (int i = 0; i < 40; i++)
      {        
        data[i].gastype = i/4 + 1;
        data[i].gasconc = i%4;
      }
      color c3, c4;
      if (data.length > 0)
      {       
        int icall = ilen / data.length;
        picturebox2.width = icall * data.length;
        picturebox1.width = icall * data.length;
        icall2 = icall * 4;
        //画对比框条
        for (int i = 0; i < 10; i++)
        {          
          brush brush1 = new lineargradientbrush(new point(0, ihigh), new point(icall2, ihigh), color[i], color[i]);
          graphic2.fillrectangle(brush1, 0 + icall2 * i, 0, icall2, ihigh);
          brush1.dispose();
        }
        //画颜色条梯度分量
        for (int i = 0; i < data.length; i++)
        {          
          //将颜色分为三个深度
          if (data[i].gasconc != 0)
            c3 = c4 = color.fromargb((byte)(255 * (float)(1 - (data[i].gasconc * 0.01))),
            (byte)(color[data[i].gastype-1].r * (float)(1 - (data[i].gasconc * 0.2))),
            (byte)(color[data[i].gastype-1].g * (float)(1 - (data[i].gasconc * 0.2))),
            (byte)(color[data[i].gastype-1].b * (float)(1 - (data[i].gasconc * 0.2))));
          else
            c3 = c4 = color.black;
          brush brush1 = new lineargradientbrush(new point(0, ihigh), new point(icall, ihigh), c3, c4);
          graphic.fillrectangle(brush1, 0 + icall * i , 0, icall, ihigh);
          brush1.dispose();                  
        }
      }
      else
      {
        c4 = color[0];
        brush brush1 = new lineargradientbrush(new point(0, ihigh), new point(ilen, ihigh), c4, c4);        
        graphic.fillrectangle(brush1, 0, 0, ilen, ihigh);
        brush1.dispose();
      }
      
    }

 public struct gasconcentration
    {
      int igastype;//气体名称
      int igasconc;//气体浓度 // 0=no, 1=low, 2=med, 3=high
 
      public int gastype { get { return igastype; }
        set { igastype = value; }    }
      public int gasconc { get { return igasconc; }
        set { igasconc = value; }
      }
    }

**这是用来获取某一颜色段的分度集合**
 /// <summary>
    /// 获得某一颜色区间的颜色集合
    /// </summary>
    /// <param name="sourcecolor">起始颜色</param>
    /// <param name="destcolor">终止颜色</param>
    /// <param name="count">分度数</param>
    /// <returns>返回颜色集合</returns>
    public static list<color> getsinglecolorlist(color srccolor, color descolor, int count)
    {
      list<color> colorfactorlist = new list<color>();
      int redspan = descolor.r - srccolor.r;
      int greenspan = descolor.g - srccolor.g;
      int bluespan = descolor.b - srccolor.b;
      for (int i = 0; i < count; i++)
      {
        color color = color.fromargb(
          srccolor.r + (int)((double)i / count * redspan),
          srccolor.g + (int)((double)i / count * greenspan),
          srccolor.b + (int)((double)i / count * bluespan)
        );
        colorfactorlist.add(color);
      }
      return colorfactorlist;
    }

**这里就是将红到紫之间的颜色分为5个区间，利用上面的算法拼接5个区间的分度值，就得到全彩颜色集合**
/// <summary>
    /// 获取从红到紫的颜色段的颜色集合
    /// </summary>
    /// <param name="totalcount">分度数</param>
    /// <param name="redtopurple">是否从红到紫色渐变</param>
    /// <returns>返回颜色集合</returns>
    public static list<color> getfullcolorlist(int totalcount, bool redtopurple = true)
    {
      list<color> colorlist = new list<color>();
      if (totalcount > 0)
      {
        if (redtopurple)
        {
          colorlist.addrange(getsinglecolorlist(color.red, color.yellow, totalcount / 5 + (totalcount % 5 > 0 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.yellow, color.lime, totalcount / 5 + (totalcount % 5 > 1 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.lime, color.cyan, totalcount / 5 + (totalcount % 5 > 2 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.cyan, color.blue, totalcount / 5 + (totalcount % 5 > 3 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.blue, color.magenta, totalcount / 5 + (totalcount % 5 > 4 ? 1 : 0)));
        }
        else
        {
          colorlist.addrange(getsinglecolorlist(color.magenta, color.blue, totalcount / 5 + (totalcount % 5 > 0 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.blue, color.cyan, totalcount / 5 + (totalcount % 5 > 1 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.cyan, color.lime, totalcount / 5 + (totalcount % 5 > 2 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.lime, color.yellow, totalcount / 5 + (totalcount % 5 > 3 ? 1 : 0)));
          colorlist.addrange(getsinglecolorlist(color.yellow, color.red, totalcount / 5 + (totalcount % 5 > 4 ? 1 : 0)));
        }
      }
      return colorlist;
    }