从原点迭代离散二维网格上的向外螺旋的算法

直接的“临时”解决方案没有错。它也可以足够干净。
只需注意螺旋是由段构建的。您可以从当前片段旋转90度获得下一个片段。并且每旋转两圈，线段的长度就会增加1。

编辑 插图，这些段编号

   ... 11 107 7 7 7 6 108 3 3 2 6 108 4 . 1 6 108 4 5 5 5 108 9 9 9 9  9

。

    // (di, dj) is a vector - direction in which we move right now    int di = 1;    int dj = 0;    // length of current segment    int segment_length = 1;    // current position (i, j) and how much of current segment we passed    int i = 0;    int j = 0;    int segment_passed = 0;    for (int k = 0; k < NUMBER_OF_POINTS; ++k) {        // make a step, add 'direction' vector (di, dj) to current position (i, j)        i += di;        j += dj;        ++segment_passed;        System.out.println(i + " " + j);        if (segment_passed == segment_length) { // done with current segment segment_passed = 0; // 'rotate' directions int buffer = di; di = -dj; dj = buffer; // increase segment length if necessary if (dj == 0) {     ++segment_length; }        }    }

要改变原来的方向，看的原始值

di

dj