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Merge pull request #1815 from juguagua/leetcode-modify-the-code-of-the-backtracking
更新回溯问题：“重新安排行程” 和 “解数独”
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程序员Carl
@ -42,9 +42,9 @@
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**如果以上这几道题目没有做过的话，不建议上来就做这道题哈！**
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[N皇后问题](https://programmercarl.com/0051.N皇后.html)是因为每一行每一列只放一个皇后，只需要一层for循环遍历一行，递归来来遍历列，然后一行一列确定皇后的唯一位置。
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[N皇后问题](https://programmercarl.com/0051.N皇后.html)是因为每一行每一列只放一个皇后，只需要一层for循环遍历一行，递归来遍历列，然后一行一列确定皇后的唯一位置。
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本题就不一样了，**本题中棋盘的每一个位置都要放一个数字（而N换后是一行只放一个皇后），并检查数字是否合法，解数独的树形结构要比N皇后更宽更深**。
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本题就不一样了，**本题中棋盘的每一个位置都要放一个数字（而N皇后是一行只放一个皇后），并检查数字是否合法，解数独的树形结构要比N皇后更宽更深**。
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因为这个树形结构太大了，我抽取一部分，如图所示：
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@ -75,7 +75,7 @@ bool backtracking(vector<vector<char>>& board)
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**那么有没有永远填不满的情况呢？**
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这个问题我在递归单层搜索逻辑里在来讲！
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这个问题我在递归单层搜索逻辑里再来讲！
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* 递归单层搜索逻辑
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@ -207,7 +207,7 @@ public:
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所以我在开篇就提到了**二维递归**，这也是我自创词汇，希望可以帮助大家理解解数独的搜索过程。
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一波分析之后，在看代码会发现其实也不难，唯一难点就是理解**二维递归**的思维逻辑。
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一波分析之后，再看代码会发现其实也不难，唯一难点就是理解**二维递归**的思维逻辑。
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**这样，解数独这么难的问题，也被我们攻克了**。
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@ -331,55 +331,56 @@ class Solution:
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### Go
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```go
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func solveSudoku(board [][]byte) {
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var backtracking func(board [][]byte) bool
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backtracking=func(board [][]byte) bool{
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for i:=0;i<9;i++{
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for j:=0;j<9;j++{
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//判断此位置是否适合填数字
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if board[i][j]!='.'{
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continue
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}
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//尝试填1-9
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for k:='1';k<='9';k++{
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if isvalid(i,j,byte(k),board)==true{//如果满足要求就填
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board[i][j]=byte(k)
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if backtracking(board)==true{
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return true
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}
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board[i][j]='.'
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}
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}
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return false
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}
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}
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return true
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}
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backtracking(board)
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func solveSudoku(board [][]byte) {
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var backtracking func(board [][]byte) bool
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backtracking = func(board [][]byte) bool {
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for i := 0; i < 9; i++ {
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for j := 0; j < 9; j++ {
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//判断此位置是否适合填数字
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if board[i][j] != '.' {
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continue
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}
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//尝试填1-9
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for k := '1'; k <= '9'; k++ {
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if isvalid(i, j, byte(k), board) == true { //如果满足要求就填
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board[i][j] = byte(k)
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if backtracking(board) == true {
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return true
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}
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board[i][j] = '.'
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}
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}
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return false
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}
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}
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return true
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}
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backtracking(board)
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}
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//判断填入数字是否满足要求
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func isvalid(row,col int,k byte,board [][]byte)bool{
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for i:=0;i<9;i++{//行
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if board[row][i]==k{
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return false
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}
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}
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for i:=0;i<9;i++{//列
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if board[i][col]==k{
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return false
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}
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}
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//方格
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startrow:=(row/3)*3
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startcol:=(col/3)*3
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for i:=startrow;i<startrow+3;i++{
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for j:=startcol;j<startcol+3;j++{
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if board[i][j]==k{
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return false
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}
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}
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}
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return true
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func isvalid(row, col int, k byte, board [][]byte) bool {
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for i := 0; i < 9; i++ { //行
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if board[row][i] == k {
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return false
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}
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}
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for i := 0; i < 9; i++ { //列
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if board[i][col] == k {
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return false
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}
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}
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//方格
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startrow := (row / 3) * 3
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startcol := (col / 3) * 3
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for i := startrow; i < startrow+3; i++ {
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for j := startcol; j < startcol+3; j++ {
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if board[i][j] == k {
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return false
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}
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}
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}
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return true
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}
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```
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@ -86,7 +86,7 @@ unordered_map<string, map<string, int>> targets：unordered_map<出发机场, ma
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在遍历 `unordered_map<出发机场, map<到达机场, 航班次数>> targets`的过程中，**可以使用"航班次数"这个字段的数字做相应的增减，来标记到达机场是否使用过了。**
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如果“航班次数”大于零，说明目的地还可以飞，如果如果“航班次数”等于零说明目的地不能飞了，而不用对集合做删除元素或者增加元素的操作。
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如果“航班次数”大于零，说明目的地还可以飞，如果“航班次数”等于零说明目的地不能飞了，而不用对集合做删除元素或者增加元素的操作。
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**相当于说我不删，我就做一个标记！**
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@ -439,68 +439,6 @@ func findItinerary(tickets [][]string) []string {
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}
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```
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### C语言
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```C
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char **result;
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bool *used;
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int g_found;
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int cmp(const void *str1, const void *str2)
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{
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const char **tmp1 = *(char**)str1;
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const char **tmp2 = *(char**)str2;
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int ret = strcmp(tmp1[0], tmp2[0]);
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if (ret == 0) {
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return strcmp(tmp1[1], tmp2[1]);
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}
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return ret;
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}
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void backtracting(char *** tickets, int ticketsSize, int* returnSize, char *start, char **result, bool *used)
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{
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if (*returnSize == ticketsSize + 1) {
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g_found = 1;
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return;
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}
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for (int i = 0; i < ticketsSize; i++) {
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if ((used[i] == false) && (strcmp(start, tickets[i][0]) == 0)) {
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result[*returnSize] = (char*)malloc(sizeof(char) * 4);
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memcpy(result[*returnSize], tickets[i][1], sizeof(char) * 4);
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(*returnSize)++;
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used[i] = true;
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/*if ((*returnSize) == ticketsSize + 1) {
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return;
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}*/
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backtracting(tickets, ticketsSize, returnSize, tickets[i][1], result, used);
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if (g_found) {
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return;
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}
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(*returnSize)--;
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used[i] = false;
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}
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}
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return;
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}
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char ** findItinerary(char *** tickets, int ticketsSize, int* ticketsColSize, int* returnSize){
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if (tickets == NULL || ticketsSize <= 0) {
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return NULL;
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}
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result = malloc(sizeof(char*) * (ticketsSize + 1));
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used = malloc(sizeof(bool) * ticketsSize);
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memset(used, false, sizeof(bool) * ticketsSize);
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result[0] = malloc(sizeof(char) * 4);
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memcpy(result[0], "JFK", sizeof(char) * 4);
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g_found = 0;
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*returnSize = 1;
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qsort(tickets, ticketsSize, sizeof(tickets[0]), cmp);
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backtracting(tickets, ticketsSize, returnSize, "JFK", result, used);
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*returnSize = ticketsSize + 1;
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return result;
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}
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```
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### Javascript
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```Javascript
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@ -589,6 +527,68 @@ function findItinerary(tickets: string[][]): string[] {
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};
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```
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### C语言
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```C
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char **result;
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bool *used;
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int g_found;
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int cmp(const void *str1, const void *str2)
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{
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const char **tmp1 = *(char**)str1;
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const char **tmp2 = *(char**)str2;
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int ret = strcmp(tmp1[0], tmp2[0]);
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if (ret == 0) {
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return strcmp(tmp1[1], tmp2[1]);
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}
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return ret;
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}
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void backtracting(char *** tickets, int ticketsSize, int* returnSize, char *start, char **result, bool *used)
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{
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if (*returnSize == ticketsSize + 1) {
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g_found = 1;
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return;
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}
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for (int i = 0; i < ticketsSize; i++) {
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if ((used[i] == false) && (strcmp(start, tickets[i][0]) == 0)) {
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result[*returnSize] = (char*)malloc(sizeof(char) * 4);
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memcpy(result[*returnSize], tickets[i][1], sizeof(char) * 4);
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(*returnSize)++;
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used[i] = true;
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/*if ((*returnSize) == ticketsSize + 1) {
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return;
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}*/
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backtracting(tickets, ticketsSize, returnSize, tickets[i][1], result, used);
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if (g_found) {
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return;
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}
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(*returnSize)--;
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used[i] = false;
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}
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}
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return;
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}
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char ** findItinerary(char *** tickets, int ticketsSize, int* ticketsColSize, int* returnSize){
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if (tickets == NULL || ticketsSize <= 0) {
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return NULL;
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}
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result = malloc(sizeof(char*) * (ticketsSize + 1));
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used = malloc(sizeof(bool) * ticketsSize);
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memset(used, false, sizeof(bool) * ticketsSize);
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result[0] = malloc(sizeof(char) * 4);
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memcpy(result[0], "JFK", sizeof(char) * 4);
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g_found = 0;
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*returnSize = 1;
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qsort(tickets, ticketsSize, sizeof(tickets[0]), cmp);
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backtracting(tickets, ticketsSize, returnSize, "JFK", result, used);
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*returnSize = ticketsSize + 1;
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return result;
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}
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```
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### Swift
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直接迭代tickets数组：
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@ -709,78 +709,6 @@ for line in tickets {
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}
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```
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### Go
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```Go
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// 先排序，然后找到第一条路径即可返回
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func findItinerary(tickets [][]string) []string {
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var path []string // 用来保存搜索的路径
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data := make(map[string]ticketSlice) // 用来保存tickets排序后的结果
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var search func(airport string) bool
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search = func(airport string) bool {
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if len(path) == len(tickets) {
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path = append(path, airport)
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return true
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}
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to := data[airport]
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for _, item := range to {
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if item.Count == 0 {
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// 已用完
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continue
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}
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path = append(path, airport)
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item.Count--
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if search(item.To) { return true }
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item.Count++
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path = path[:len(path) - 1]
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}
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return false
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}
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// 排序
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// 感觉这段代码有点啰嗦，不知道能不能简化一下
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tmp := make(map[string]map[string]int)
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for _, ticket := range tickets {
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if to, ok := tmp[ticket[0]]; ok {
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if _, ok2 := to[ticket[1]]; ok2 {
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to[ticket[1]]++
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} else {
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to[ticket[1]] = 1
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}
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} else {
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tmp[ticket[0]] = map[string]int{
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ticket[1]: 1,
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}
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}
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}
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for from, to := range tmp {
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var tmp ticketSlice
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for to, num := range to {
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tmp = append(tmp, &ticketStat{To: to, Count: num})
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}
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sort.Sort(tmp)
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data[from] = tmp
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}
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search("JFK")
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return path
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}
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type ticketStat struct {
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To string
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Count int
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}
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type ticketSlice []*ticketStat
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func (p ticketSlice) Len() int { return len(p) }
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func (p ticketSlice) Less(i, j int) bool { return strings.Compare(p[i].To, p[j].To) == -1 }
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func (p ticketSlice) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
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```
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### Rust
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** 文中的Hashmap嵌套Hashmap的方法因为Rust的所有权问题暂时无法实现，此方法为删除哈希表中元素法 **
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```Rust
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