Day27题目

LeetCode39组合总和

核心思想：经典回溯

class Solution {int sum = 0;//记录当前元素List<Integer> path = new ArrayList<>();// 记录返回值List<List<Integer>> res = new ArrayList<>();public List<List<Integer>> combinationSum(int[] candidates, int target) {help(candidates,0,target);return res;}public void help(int[] candidates,int start,int target){// 返回条件是sum 大于target，这样就不能再有和等于target，因为元素全部是正数if(sum > target) return;for(int i = start ; i < candidates.length; i ++){sum = sum + candidates[i];path.add(candidates[i]);if(sum == target){res.add(new ArrayList(path));}else{help(candidates,i,target);}// 回溯sum = sum-candidates[i];path.removeLast();}}
}

LeetCode40组合总和Ⅱ：多了去重

核心思想：主要是去重部分：去重有好几种方法：使用Boolean数组，set，排序后比较是否和前一个元素相同

class Solution {List<Integer> path = new ArrayList<>();List<List<Integer>> res = new ArrayList<>();public List<List<Integer>> combinationSum2(int[] candidates, int target) {// 不使用used 数组Arrays.sort(candidates);help(candidates,target,0,0);return res;}public void help(int[] candidates, int target, int start, int sum){if(sum > target) return;if(sum == target) {res.add(new ArrayList(path));}// 这里做了剪枝  4ms->2msfor(int i = start; i < candidates.length && sum+candidates[i] <= target ;i ++){// 这里是在层部分判断是否是和前一个相同。 i>start 则表示是同一层if(i > start && candidates[i-1] == candidates[i]) continue;path.add(candidates[i]);help(candidates,target,i+1,sum+candidates[i]);path.removeLast();}}
}

LeetCode131.分割回文串

核心思想：使用start维护一个字符串的起始

class Solution {List<String> path = new ArrayList<>();List<List<String>> res = new ArrayList<>();public List<List<String>> partition(String s) {help(s,0);return res;}public void help(String s,int start){// 左闭右开if(start >= s.length()){res.add(new ArrayList(path));return;}// 从start 开始，可以切 1 -> s.length-start-1 种截取的方法。for(int i = start+1; i <= s.length();  i++ ){if(isReverse(s,start,i)){// 是回文串之后进行进一步遍历path.add(s.substring(start,i));help(s,i);path.removeLast();}}}public boolean isReverse(String s,int start,int end){for (int i = start, j = end-1; i < j; i++, j--) {if (s.charAt(i) != s.charAt(j)) {return false;}}return true;}
}