组合:
public class Solution { static int count = 0; static void findSo(String soFar,String rest) { if(soFar.length() == 3){ count++; System.out.println(soFar); }else{ for(int i=0; i<rest.length(); i++){ String now; if(soFar.length()<3){ now = soFar+rest.charAt(i); }else{ now = soFar; } String remain = rest.substring(i+1); findSo(now,remain); } } } public static void main(String[] args) { String str ="1234567"; findSo("",str); System.out.println("count:"+count); } }
结果:
123 124 125 126 127 134 135 136 137 145 146 147 156 157 167 234 235 236 237 245 246 247 256 257 267 345 346 347 356 357 367 456 457 467 567 count:35
C++版本
#include <iostream> #include <string> #include <stdio.h> using namespace std; int count; void findSo(string soFar, string rest) { if (soFar.length() == 3) { count++; cout<<soFar<<endl; } else { for (int i = 0; i < rest.length(); i++) { string now; if(soFar.length()<3){ now = soFar + rest.at(i); }else{ now = soFar; } string remain = rest.substr(i + 1); findSo(now, remain); } } } int main() { string str = "1234567"; findSo("", str); cout<<"count:"<<count<<endl; }
排列:
public class Solution { static int count = 0; static void findSo(String soFar, String rest) { if (soFar.length() == 3) { count++; System.out.println(soFar); } else { for (int i = 0; i < rest.length(); i++) { String now = soFar + rest.charAt(i); String remain = rest.substring(0,i) + rest.substring(i + 1); findSo(now, remain); } } } public static void main(String[] args) { String str = "1234567"; findSo("", str); System.out.println("count:" + count); } }
for循环里面的代码也可以和组合一样
for (int i = 0; i < rest.length(); i++) { String now; if (soFar.length() < 3) { now = soFar + rest.charAt(i); } else { now = soFar; } String remain = rest.substring(0, i) + rest.substring(i + 1); findSo(now, remain); }
结果:
123 124 125 126 127 132 134 135 136 137 142 143 145 146 147 152 153 154 156 157 162 163 164 165 167 172 173 174 175 176 213 214 215 216 217 231 234 235 236 237 241 243 245 246 247 251 253 254 256 257 261 263 264 265 267 271 273 274 275 276 312 314 315 316 317 321 324 325 326 327 341 342 345 346 347 351 352 354 356 357 361 362 364 365 367 371 372 374 375 376 412 413 415 416 417 421 423 425 426 427 431 432 435 436 437 451 452 453 456 457 461 462 463 465 467 471 472 473 475 476 512 513 514 516 517 521 523 524 526 527 531 532 534 536 537 541 542 543 546 547 561 562 563 564 567 571 572 573 574 576 612 613 614 615 617 621 623 624 625 627 631 632 634 635 637 641 642 643 645 647 651 652 653 654 657 671 672 673 674 675 712 713 714 715 716 721 723 724 725 726 731 732 734 735 736 741 742 743 745 746 751 752 753 754 756 761 762 763 764 765 count:210
C++版本:
#include <iostream> #include <string> #include <stdio.h> using namespace std; int count; void findSo(string soFar, string rest) { if (soFar.length() == 3) { count++; cout<<soFar<<endl; } else { for (int i = 0; i < rest.length(); i++) { string now = soFar + rest.at(i); string remain = rest.substr(0,i) + rest.substr(i + 1); findSo(now, remain); } } } int main() { string str = "1234567"; findSo("", str); cout<<"count:"<<count<<endl; }
从 m个元素中去n个元素的排列和组合差别仅仅在于
排列 string remain = rest.substr(0,i)+rest.substr(i + 1);
组合 string remain = rest.substr(i + 1);
所以组合代码也可以是:
#include <iostream> #include <string> #include <stdio.h> using namespace std; int count; void findSo(string soFar, string rest) { if (soFar.length() == 3) { count++; cout<<soFar<<endl; } else { for (int i = 0; i < rest.length(); i++) { string now = soFar + rest.at(i); string remain = rest.substr(i + 1); findSo(now, remain); } } } int main() { string str = "1234567"; findSo("", str); cout<<"count:"<<count<<endl; }
ps:
抽象的想一下 以1 2 3 4 5 6 7为例
for循环刚进来 递归肯定会得到7个 大分支
不好意思,他现在先要处理第一个分支
即("1","2,3,4,5,6,7) 在这个分支下不断的分
递归每得到一个分支就要穷尽这个分支
穷尽之后在返回到父节点的相邻分支 继续穷尽
for循环刚进来 递归肯定会得到7个 大分支
不好意思,他现在先要处理第一个分支
即("1","2,3,4,5,6,7) 在这个分支下不断的分
递归每得到一个分支就要穷尽这个分支
穷尽之后在返回到父节点的相邻分支 继续穷尽
