leetCode解题报告5道题(十一) - c++编程基础

题目一：Subsets

Given a set of distinct integers, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.

The solution set must not contain duplicate subsets.

For example,
If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

分析：

题目的题意挺简单的，我就不做说明，主要这道题目可以用递归的方法来做，也可以直接用迭代来取代，我们用迭代的方式来做这道题目，没有什么特别的算法，直接上代码，代码里有注释哈！

AC代码：

public class Solution {
    private ArrayList
    
     > results = new ArrayList
     
      >(); public ArrayList
      
       > subsets(int[] S) { int len = S.length; results.add(new ArrayList
       
        ()); //先增加一个 [] 空集 int size = results.size(); int nowDealCount = 0; //表示当前要处理的长度 while (nowDealCount != len){ for (int i=0; i
        
          nowList = (ArrayList
         
          )results.get(i); if (nowDealCount == 0){//如果处理的长度为0,表示是最开始只有一个 空集 的时候 for (int j=0; j
          
            addList = new ArrayList
           
            (); addList.add(S[j]); results.add(addList); } }else if (nowList != null && nowList.size() == nowDealCount){ int maxNum = (int)nowList.get(nowDealCount-1); for (int j=0; j
            
              copyList = new ArrayList
             
              (nowList); copyList.add(S[j]); results.add(copyList); } } } } size = results.size(); nowDealCount++; } return results; } }

题目二：Subsets II

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:
```
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]
```
分析：

这道题目跟上面那道基本上思路是类似的，只是这道题目多了可能重复的元素，这时候我们就要考虑重复的元素不能有相同集合的问题了，先把Num[]数组进行排序，然后跟题目一一样的解法，只是在此过程中我们需要知道每个添加到results中的集合的最后一个元素(即最大元素)的下标，我们用一个indexList来存储。

这样的话每次要再添加新元素组成新的集合，就只要从下标位置+1开始，到len结束来搜索就行了。但要注意的是，如果当前要添加到集合中的元素和前一个元素相等，那么就不添加，如： [1, 2, 2], 当我们处理到nowDealCount==1时

要往 [1] 中添加元素，我们会添加一个 [2]，组成[1, 2]但是当到第二个[2]的时候，我们不能再添加一个[1, 2]了，否则就重复出现了！

AC代码：
```
public class Solution {
    private ArrayList
      
       > results = new ArrayList
       
        >(); public ArrayList
        
         > subsets(int[] S) { int len = S.length; results.add(new ArrayList
         
          ()); //先增加一个 [] 空集 int size = results.size(); int nowDealCount = 0; //表示当前要处理的长度 while (nowDealCount != len){ for (int i=0; i
          
            nowList = (ArrayList
           
            )results.get(i); if (nowDealCount == 0){//如果处理的长度为0,表示是最开始只有一个 空集 的时候 for (int j=0; j
            
              addList = new ArrayList
             
              (); addList.add(S[j]); results.add(addList); } }else if (nowList != null && nowList.size() == nowDealCount){ int maxNum = (int)nowList.get(nowDealCount-1); for (int j=0; j
              
                copyList = new ArrayList
               
                (nowList); copyList.add(S[j]); results.add(copyList); } } } } size = results.size(); nowDealCount++; } return results; } }
               
              
             
            
           
          
         
        
       
      
```
题目三：Unique Paths

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there

Note: m and n will be at most 100.

分析：

题目的意思从一个矩阵的左上角的位置开始走，只能向右走或者向下走，最后要走到矩阵的右下角，求解出一共有多少种不同的走法.

我们拿一个示例来分析下，当m=3, n=7的时候

一个3 * 7的矩阵

vc/Cw+a1xMq919M8L3A+CjxwPjxicj4KPC9wPgo8cD48c3Ryb25nPndheXNbbS0xXVtjb2xdID0gMTsgKCAwIDw9IGNvbCA8IG4pPC9zdHJvbmc+PC9wPgo8cD48c3Ryb25nPndheXNbcm93XVtuLTFdID0gMTsgKDAgPD0gcm93IDwgbSk8L3N0cm9uZz48L3A+CjxwPjxzdHJvbmc+d2F5c1tyb3ddW2NvbF0gPSB3YXlzW3JvdyYjNDM7MV1bY29sXSAmIzQzOyB3YXlzW3Jvd11bY29sJiM0MzsxXSAgKDA8PSByb3cgPD1tLTIsIDA8PSBjb2wgPD1uL

leetCode解题报告5道题(十一)(一)

题目一：Subsets

题目二：Subsets II

题目三：Unique Paths