leetcode题解之39. 组合总和

• Java
• Python
• C++

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.List;
public class Solution {
<span class="hljs-keyword">public</span> List&lt;List&lt;Integer&gt;&gt; combinationSum(<span class="hljs-keyword">int</span>[] candidates, <span class="hljs-keyword">int</span> target) {
    List&lt;List&lt;Integer&gt;&gt; res = <span class="hljs-keyword">new</span> ArrayList&lt;&gt;();
    <span class="hljs-keyword">int</span> len = candidates.length;

    <span class="hljs-comment">// 排序是为了提前终止搜索</span>
    Arrays.sort(candidates);

    dfs(candidates, len, target, <span class="hljs-number">0</span>, <span class="hljs-keyword">new</span> ArrayDeque&lt;&gt;(), res);
    <span class="hljs-keyword">return</span> res;
}

<span class="hljs-comment">/**
 * <span class="hljs-doctag">@param</span> candidates 数组输入
 * <span class="hljs-doctag">@param</span> len        输入数组的长度，冗余变量
 * <span class="hljs-doctag">@param</span> residue    剩余数值
 * <span class="hljs-doctag">@param</span> begin      本轮搜索的起点下标
 * <span class="hljs-doctag">@param</span> path       从根结点到任意结点的路径
 * <span class="hljs-doctag">@param</span> res        结果集变量
 */</span>
<span class="hljs-function"><span class="hljs-keyword">private</span> <span class="hljs-keyword">void</span> <span class="hljs-title">dfs</span><span class="hljs-params">(<span class="hljs-keyword">int</span>[] candidates,
                 <span class="hljs-keyword">int</span> len,
                 <span class="hljs-keyword">int</span> residue,
                 <span class="hljs-keyword">int</span> begin,
                 Deque&lt;Integer&gt; path,
                 List&lt;List&lt;Integer&gt;&gt; res)</span> </span>{
    <span class="hljs-keyword">if</span> (residue == <span class="hljs-number">0</span>) {
        <span class="hljs-comment">// 由于 path 全局只使用一份，到叶子结点的时候需要做一个拷贝</span>
        res.add(<span class="hljs-keyword">new</span> ArrayList&lt;&gt;(path));
        <span class="hljs-keyword">return</span>;
    }

    <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = begin; i &lt; len; i++) {

        <span class="hljs-comment">// 在数组有序的前提下，剪枝</span>
        <span class="hljs-keyword">if</span> (residue - candidates[i] &lt; <span class="hljs-number">0</span>) {
            <span class="hljs-keyword">break</span>;
        }

        path.addLast(candidates[i]);
        dfs(candidates, len, residue - candidates[i], i, path, res);
        path.removeLast();

    }
}

}

from typing import List

class Solution:

def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:

size = len(candidates)

if size == 0:

return []
    <span class="hljs-comment"># 剪枝是为了提速，在本题非必需</span>
    candidates.sort()
    <span class="hljs-comment"># 在遍历的过程中记录路径，它是一个栈</span>
    path = []
    res = []
    <span class="hljs-comment"># 注意要传入 size ，在 range 中， size 取不到</span>
    self.__dfs(candidates, <span class="hljs-number">0</span>, size, path, res, target)
    <span class="hljs-keyword">return</span> res

<span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">__dfs</span><span class="hljs-params">(self, candidates, begin, size, path, res, target)</span>:</span>
    <span class="hljs-comment"># 先写递归终止的情况</span>
    <span class="hljs-keyword">if</span> target == <span class="hljs-number">0</span>:
        <span class="hljs-comment"># Python 中可变对象是引用传递，因此需要将当前 path 里的值拷贝出来</span>
        <span class="hljs-comment"># 或者使用 path.copy()</span>
        res.append(path[:])
        <span class="hljs-keyword">return</span>

    <span class="hljs-keyword">for</span> index <span class="hljs-keyword">in</span> range(begin, size):
        residue = target - candidates[index]
        <span class="hljs-comment"># “剪枝”操作，不必递归到下一层，并且后面的分支也不必执行</span>
        <span class="hljs-keyword">if</span> residue &lt; <span class="hljs-number">0</span>:
            <span class="hljs-keyword">break</span>
        path.append(candidates[index])
        <span class="hljs-comment"># 因为下一层不能比上一层还小，起始索引还从 index 开始</span>
        self.__dfs(candidates, index, size, path, res, residue)
        path.pop()

if name == ‘main‘:

candidates = [2, 3, 6, 7]

target = 7

solution = Solution()

result = solution.combinationSum(candidates, target)

print(result)

// author:rmokerone

#include <iostream>

#include <vector>

using namespace std;
class Solution {

private:

vector<int> candidates;

vector<vector<int>> res;

vector<int> path;

public:

void DFS(int start, int target) {

if (target == 0) {

res.push_back(path);

return;

}

for (int i = start;

i < candidates.size() && target - candidates[i] >= 0; i++) {

path.push_back(candidates[i]);

DFS(i, target - candidates[i]);

path.pop_back();

}

}
<span class="hljs-function"><span class="hljs-built_in">vector</span>&lt;<span class="hljs-built_in">vector</span>&lt;<span class="hljs-keyword">int</span>&gt;&gt; <span class="hljs-title">combinationSum</span><span class="hljs-params">(<span class="hljs-built_in">vector</span>&lt;<span class="hljs-keyword">int</span>&gt; &amp;candidates, <span class="hljs-keyword">int</span> target)</span> </span>{
    <span class="hljs-built_in">std</span>::sort(candidates.begin(), candidates.end());
    <span class="hljs-keyword">this</span>-&gt;candidates = candidates;
    DFS(<span class="hljs-number">0</span>, target);

    <span class="hljs-keyword">return</span> res;
}

};