DFS习题复习

 

DFS(深度优先搜索)解题思路：recursion的思想。（常见题型：当题目要求穷举所有可能性时，多用DFS），解题思路分两步：

1：考虑总共recursion有几层

2：考虑每一层recursion内有多少个case

1.All Subsets

Given a set of characters represented by a String, return a list containing all subsets of the characters.

Assumptions

• There are no duplicate characters in the original set.

​Examples

• Set = "abc", all the subsets are [“”, “a”, “ab”, “abc”, “ac”, “b”, “bc”, “c”]
• Set = "", all the subsets are [""]
• Set = null, all the subsets are []

解题思路：1：recursion层数为：set.length()

               2:每层case，只需考虑当前layer对应的character存在或不存在

recursion tree 为(以“abc”为例)：                 {}

                     "a" level:                   {a}                 {}

                     "b"level:            {ab}        {a}      {b}       {}  

                     "c"level:       {abc}  {ab} {ac} {a} {bc} {b} {c} {}

由recursion tree可以看出，当递归进行到最后一层时，所有得leaf node就为所有得结果：

Java Solution:

public List<String> subSets(String set) {
   List<String>result=new ArrayList<String>();
   if(set == null){
     return result;
   }
   helper(set, new StringBuilder(), 0, result);
   return result;
  }
  private void helper(String set, StringBuilder sb, int layer, List<String>result){
    if(layer == set.length()){
      result.add(sb.toString());
      return;
    }
    sb.append(set.charAt(layer));
    helper(set, sb, layer+1, result);
    sb.deleteCharAt(sb.length()-1);
    helper(set, sb, layer+1, result);
  }

考虑存在重复元素的情况：

去重方法：

Solution1：如果数组是已经排序好的，我们可以比较容易的去处重复元素，在每一层layer进行完后，只需要跳到下一个和当前值不同的节点即可。

由于整个DFS的时间复杂度为O（2^n），因此可以先将传入数组排序后再进行DFS过程，由于排序时间复杂度仅为o（nlogn）因此并不影响整体的时间复杂度。

Solution2：如果不进行排序，也可以每层maintain一个hashset，存放之前层出现过的charater，跳过所有之前存在过的节点即可。

public List<String> subSets(String set) {
        List<String> result = new ArrayList<String>();
        if (set == null) {
            return result;
        }
        if(set.length() == 0) {
            result.add("");
            return result;
        }
        char[]array=set.toCharArray();
        Arrays.sort(array);
        set=new String(array);
        helper(result,set,0,new StringBuilder());
        return result;
    }
    private void helper(List<String>result,String set, int layer, StringBuilder sb) 
       {
        if(layer == set.length()) {
            result.add(sb.toString());
            return;
        }
        sb.append(set.charAt(layer));
        helper(result,set,layer+1,sb);
        sb.deleteCharAt(sb.length()-1);
        while(layer < set.length()-1 && set.charAt(layer+1) == set.charAt(layer)) {
            layer++;
        }
        helper(result,set,layer+1,sb);
    }

.

2.All Permutations:

Given a string with no duplicate characters, return a list with all permutations of the characters.

Examples

• Set = “abc”, all permutations are [“abc”, “acb”, “bac”, “bca”, “cab”, “cba”]
• Set = "", all permutations are [""]
• Set = null, all permutations are []

解题思路：

1：总共层数，layer=set.length();

2：每层case：同subsets的区别在于，每个character仅是位置互换，而非考虑其存在不存在。因此在每一层recursion中，我们只需穷举所有可能在当前位置出现的character的可能性。具体的实现方法，将当前节点后的每个节点的值和当前节点swap，递归完后再swap回来再交换下一个节点，简称“swap来swap去”

PS:若题目要求需要考虑有重复元素的情况时，去重方法与上题完全相同，既可以从sort的角度去考虑，也可以用Hashset的方法。

Java Solution：

public List<String> permutations(String set) {
  List<String>result=new ArrayList<String>();
  if(set==null){
    return result;
  }
  char[]c=set.toCharArray();
  helper(c, result, 0);
  return result;
  }
  private void helper(char[]c, List<String>result,int index){
    if(index == c.length){
      result.add(new String(c));
      return;
    }
    for(int i=index; i<c.length; i++){
      swap(c, i, index);
      helper(c, result, index+1);
      swap(c, i, index);
    }
  }
  private void swap(char[]array, int i, int j){
    char temp=array[i];
    array[i]=array[j];
    array[j]=temp;
  }

3：All Valid Permutations Of Parentheses

Given N pairs of parentheses “()”, return a list with all the valid permutations.

Assumptions

• N >= 0

Examples

• N = 1, all valid permutations are ["()"]
• N = 3, all valid permutations are ["((()))", "(()())", "(())()", "()(())", "()()()"]
• N = 0, all valid permutations are [""]

解题思路：

1：总共有多少层？ 总共层数为括号总数：n*2

2：每层有多少case？ 与之前两题不同的是，本题需要考虑左括号和右括号添加priority的问题，既必须得保证，在每次添加右括号时，得保证之前添加的左括号个数大于右括号个数。既如果之前添加过的括号为“（）”，在当前层仅能加左括号而不能加右括号，因此每层的case分为两个，既要么加左括号，要么加右括号。

Java Solution:

public List<String> validParentheses(int n) {
   List<String>result=new ArrayList<String>();
   helper(n, 0, 0, result, new StringBuilder());
   return result;
  } 
  private void helper(int n, int left, int right, List<String>result, StringBuilder sb){
    if(left == n && right == n){
      result.add(sb.toString());
      return;
    }
    if(left < n){
      sb.append("(");
      helper(n, left+1, right, result, sb);
      sb.deleteCharAt(sb.length()-1);
    }
    if(right < left){
      sb.append(")");
      helper(n, left, right+1, result, sb);
      sb.deleteCharAt(sb.length()-1);
    }
  }

4:Combination of Coins

Given a number of different denominations of coins (e.g., 1 cent, 5 cents, 10 cents, 25 cents), get all the possible ways to pay a target number of cents.

Arguments

• coins - an array of positive integers representing the different denominations of coins, there are no duplicate numbers and the numbers are sorted by descending order, eg. {25, 10, 5, 2, 1}
• target - a non-negative integer representing the target number of cents, eg. 99

Assumptions

• coins is not null and is not empty, all the numbers in coins are positive
• target >= 0
• You have infinite number of coins for each of the denominations, you can pick any number of the coins.

Return

• a list of ways of combinations of coins to sum up to be target.
• each way of combinations is represented by list of integer, the number at each index means the number of coins used for the denomination at corresponding index.

Examples

coins = {2, 1}, target = 4, the return should be

[

  [0, 4],   (4 cents can be conducted by 0 * 2 cents + 4 * 1 cents)

  [1, 2],   (4 cents can be conducted by 1 * 2 cents + 2 * 1 cents)

  [2, 0]    (4 cents can be conducted by 2 * 2 cents + 0 * 1 cents)

]

解题思路：

1：多少层？： coins.length 层（硬币种类数）每一层仅需考虑当前种类的硬币可能存在多少种情况

2：每层多少case？：每层case为每种硬币在当前target的条件下，所有可能出现的情况。比如： 当前剩余硬币为100， 当前硬币面值为3，那当前面值为3的硬币的所有可能性为0-100/3 次，分析时间复杂度时，取worst case，既面值最小硬币的情况，如target=100，最小硬币面值为1时，时间复杂度为o（100^coins.length）

Java Solution:

public List<List<Integer>> combinations(int target, int[] coins){
     List<List<Integer>>result=new ArrayList<List<Integer>>();
   helper(target, coins, result, new ArrayList<Integer>(), 0);
   return result;
    }
    private void helper(int target, int[]coins, List<List<Integer>>result,
    List<Integer>list, int layer){
      if(layer == coins.length-1){
        if(target % coins[layer] == 0){
          list.add(target/coins[layer]);
          result.add(new ArrayList(list));
          list.remove(list.size()-1);
        }
        return;
      }
      for(int i=0; i <= target/coins[layer]; i++){
        list.add(i);
        helper(target-(i*coins[layer]), coins, result, list, layer+1);
        list.remove(list.size()-1);
      }
    }

5: N皇后问题

Get all valid ways of putting N Queens on an N * N chessboard so that no two Queens threaten each other.

Assumptions

• N > 0

Return

• A list of ways of putting the N Queens
• Each way is represented by a list of the Queen's y index for x indices of 0 to (N - 1)

Example

N = 4, there are two ways of putting 4 queens:

[1, 3, 0, 2] --> the Queen on the first row is at y index 1, the Queen on the second row is at y index 3, the Queen on the third row is at y index 0 and the Queen on the fourth row is at y index 2.

[2, 0, 3, 1] --> the Queen on the first row is at y index 2, the Queen on the second row is at y index 0, the Queen on the third row is at y index 3 and the Queen on the fourth row is at y index 1.

 解题思路与之前题类似，还是分析两个条件：

1：总共有多少层？：显然有N层。

2：每层有多少个case？ 每层理论上来说，需要考虑每一个位置，但是并不是所有的节点都valid，由于上下及斜线不能有重复节点，所以在每层递归到下一层前需要先判断该位置是否valid，时间复杂度为n！.

Java Solution：

 

public List<List<Integer>> nqueens(int n) {
    List<List<Integer>>result=new ArrayList<List<Integer>>();
    helper(result, new ArrayList<Integer>(), n, 0);
    return result;
  }
  private void helper(List<List<Integer>>result, List<Integer>list, int n, int layer){
    if(n == layer){
      result.add(new ArrayList<Integer>(list));
      return;
    }
    for(int i=0; i<n; i++){
      if(isValid(i, list)){
        list.add(i);
        helper(result, list, n, layer+1);
        list.remove(list.size()-1);
      }
    }
  }
  private boolean isValid(int i, List<Integer>list){
    if(list.isEmpty()){
      return true;
    }
    int layer=0;
    for(Integer in: list){
      if(i == in || Math.abs(i-in) == Math.abs(list.size()-layer)){
        return false;
      }
      layer++;
    }
    return true;
  }