0001 void buildSS(char* P, int m, int* ss) { //O(m = |P|)
0002    ss[m-1] = m; //确定末字符对应的项；以下，自后向前地扫描P，依次计算出其余各项
0003    for (int a = m-2, b = m-2, d = 1, j = m-2; j >= 0; j--) //d = m-b-1, (a,b] = (a+d,m)
0004       if (ss[j+d] < j-a) //情况一：虽自然有[j-ss[j+d],j] = [j+d-ss[j+d],j+d]
0005          ss[j] = ss[j+d]; //但[j+d-ss[j+d]] <> [m-1-ss[j+d]]，故ss[j] <= ss[j+d]
0006       else { //情况二：已有(a,j] = (a+d,j+d]，MS[j]的左边界还有可能进一步拓展
0007          a = min(a,j);  b = j;  d = m-b-1; //更新(a,b]与d
0008          while ((0 <= a) && (P[a] == P[a+d])) //逐个比对，向左拓展(a,b]
0009             a--; //所有的while循环均由a控制，而a全程从m-2到0单调递减，故累计仅O(m)
0010          ss[j] = b-a; //MS[j] = 拓展之后的(a,b]
0011       } //else
0012 } //buildSS