leetcode题目链接
39. 组合总和
40.组合总和II
131.分割回文串
- 组合总和
class Solution {List<Integer> path = new ArrayList();List<List<Integer>> result = new ArrayList();public List<List<Integer>> combinationSum(int[] candidates, int target) {result.clear();path.clear();Arrays.sort(candidates);backTrace(candidates,target,0,0);return result;}public void backTrace(int[] candidates,int target,int startIndex,int sum){//确定终止条件//if(sum > target) return;if(sum == target){result.add(new ArrayList(path));return;}//横向的是遍历//sum+candidates[i] <target//此处剪枝操作需要先给数组排序，如果下一层的sum（就是本层的 sum + candidates[i]）//已经大于target，就可以结束本轮for循环的遍历。for(int i = startIndex;i<candidates.length && sum+candidates[i] <=target;i++){sum += candidates[i];path.add(candidates[i]);//树往下递归的时候，可以使用上一个节点的元素ibackTrace(candidates,target,i,sum);sum -= candidates[i];path.removeLast();}}
}
40.组合总和II
class Solution {List<Integer> path = new ArrayList();List<List<Integer>> result = new ArrayList();boolean[] used =null;public List<List<Integer>> combinationSum2(int[] candidates, int target) {used = new boolean[candidates.length];Arrays.fill(used,false);Arrays.sort(candidates);backTrace(candidates,target,0,0,used);return result;}public void backTrace(int[] candidates,int target,int startIndex,int sum,boolean[] used){//确定终止条件if(sum > target) return;if(sum == target){result.add(new ArrayList(path));return;}for(int i = startIndex;i<candidates.length && candidates[i] + sum <= target;i++){//树层去重，树枝不去重if(i>0 && candidates[i-1] == candidates[i] && !used[i-1])continue;path.add(candidates[i]);sum += candidates[i];used[i] = true;backTrace(candidates,target,i+1,sum,used);path.removeLast();used[i] = false;sum -=candidates[i];}}
}
131.分割回文串
class Solution {List<List<String>> result = new ArrayList();List<String> path = new ArrayList();public List<List<String>> partition(String s) {backTrace(s,0);return result;}public void backTrace(String s,int startIndex){//结束条件是截取到最后if(startIndex == s.length()){result.add(new ArrayList(path));return;}for(int i = startIndex;i<s.length();i++){if(isPalindrome(s,startIndex,i)){String str = s.substring(startIndex,i+1);path.add(str);}else continue;backTrace(s,i+1);path.removeLast();}}//判断回文数public boolean isPalindrome(String s,int start,int end){for(int i = start, j=end;i<j;i++,j--){if(s.charAt(i) != s.charAt(j)){return false;}}return true;}
}
)
)
