代码随想录算法训练营第二十五天|216.组合总和III、17.电话号码的字母组合
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216.组合总和III
- 刷题https://leetcode.cn/problems/combination-sum-iii/description/
- 文章讲解https://programmercarl.com/0216.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8CIII.html#%E7%AE%97%E6%B3%95%E5%85%AC%E5%BC%80%E8%AF%BE
- 视频讲解https://www.bilibili.com/video/BV1wg411873x/?vd_source=af4853e80f89e28094a5fe1e220d9062
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回溯树图解:
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题解:
class Solution {//递归List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combinationSum3(int k, int n) {//n为目标总和,k为树高度,即选取元素个数combinationSum31(n, k, 1, 0);return result;}//targetSum为目标总和,sum为当前总和//k为数的个数,startIndex为每层的每个结点的开始Indexpublic void combinationSum31(int targetSum, int k, int startIndex, int sum){//剪枝1:当目前sum超过目标sum时,剪枝if(sum > targetSum){return;}//出口//当path长度增加到了k,再长则无意义,结束,返回if(path.size() == k){//当长度到了k且sum符合要求,则将当前序列加入结果集,返回//若长度达到k但sum不符合要求,就直接返回if(sum == targetSum){result.add(new ArrayList<>(path));}return;}//递归回溯部分//剪枝2:控制层序遍历范围//for循环横向遍历过程(只能使用数字1-9)for(int i = startIndex; i <= 9 - (k - path.size()) + 1; i++){path.add(i);sum += i;//递归纵向遍历过程combinationSum31(targetSum, k, i + 1, sum);//回溯部分path.removeLast();sum -= i;}}
}
17.电话号码的字母组合
- 刷题https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/
- 文章讲解https://programmercarl.com/0017.%E7%94%B5%E8%AF%9D%E5%8F%B7%E7%A0%81%E7%9A%84%E5%AD%97%E6%AF%8D%E7%BB%84%E5%90%88.html
- 文章讲解https://programmercarl.com/0017.%E7%94%B5%E8%AF%9D%E5%8F%B7%E7%A0%81%E7%9A%84%E5%AD%97%E6%AF%8D%E7%BB%84%E5%90%88.html
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回溯树图解:
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题解:
class Solution {//全局列表存储最后结果List<String> list = new ArrayList<>();//与组合总和题目区别在于://此处为在两个集合中选择,所以也不需要startIndex来剪枝避免重复序列了public List<String> letterCombinations(String digits) {if(digits == null || digits.length() == 0){return list;}//二维数组设置数字字母对应关系String[] numString = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};letterCombinations1(digits, numString, 0);return list;}//因为涉及大量字符串拼接,选用StringBuilderStringBuilder temp = new StringBuilder();//num:个数最多num个public void letterCombinations1(String digits, String[] numString, int num){//递归出口,当前字符串长度达到要求,则加入结果集并返回if(num == digits.length()){list.add(temp.toString());return;}//str记录当前num对应的字符串(先字符转化为数字,再取出数字对应的字母字符串)String str = numString[digits.charAt(num) - '0'];for(int i = 0; i < str.length(); i++){temp.append(str.charAt(i));//递归letterCombinations1(digits, numString, num + 1);//回溯(弹出最后一个)temp.deleteCharAt(temp.length() - 1);}}
}