Huawei machine exam questions

Reverse placement of a single-chain table

Void sor (node ** head) {If (! Head) return; // node * cur = * head; If (! Cur |! Cur-> next) return; // The list is empty, or there is only one node // cur // | // v // node1-> node2-> node3->... node5node * Prev = NULL; while (cur & cur-> next) {// save curnode * temp = cur; // cur, temp // | // v // node1-> node2-> node3->... node5 // move a cur = cur-> next; // temp cur // | // v // node1-> node2-> node3->... node5 // The next pointer of the temporary node points to the previous prevtemp-> next = Prev; // temp cur // | // v // node1 node2-> node3->... node5 // | // v // Prev (null) // give the pointer value of the temporary node to prevprev = temp; // temp, prev cur // | // v // node1 node2-> node3->... node5 // | // v // (null)} cur-> next = Prev; * head = cur ;}

Integer to String Conversion

Void change (INT number, char s []) {bool is_minus = (number <0); number = is_minus? Number *-1: Number; char * x = s; do {* x ++ = Number % 10 + '0'; number/= 10 ;}while (number ); if (is_minus) * x ++ = '-'; // Terminator * x = '\ 0'; // invert int Len = strlen (s ); for (INT I = 0; I <(len-1)/2; I ++) {char c = s [I]; s [I] = s [len-i-1]; s [len-i-1] = C;} return ;}

Maximum text return

Void Huiwen (char input [], int Len, char output []) {for (INT I = Len; I> = 1; I --) {for (Int J = 0; j <= len-I; j ++) {// judge input [j... (len-i-1-j)] whether this paragraph is a text back // use the classic retrieval algorithm int head = J; int tail = J + I-1; while (Head <tail & input [head ++] = input [tail --]); // subscripts are staggered // If (head> = tail) {// memcpy (output, input + J, I); Return ;}}} is returned after I is copied from J ;}}}}