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开幕雷击,感觉挺难的,需要把切割问题抽象为组合问题,还得设置切割过的地方不能重复切割,所以递归函数需要传入i + 1。
模拟切割线,其实就是index是上一层已经确定了的分割线,i是这一层试图寻找的新分割线
- class Solution {
- List<List<String>> res = new ArrayList<>();
- List<String> path = new LinkedList<>();
- public List<List<String>> partition(String s) {
- backtrack(s,0);
- return res;
- }
- void backtrack(String s, int start) {
- if (start >= s.length()) {
- res.add(new ArrayList(path));
- return;
- }
- for (int i = start; i < s.length(); i++) {
- if (isPalindrome(s, start, i)) {
- String str = s.substring(start, i + 1);
- path.addLast(str);
- } else {
- continue;
- }
- backtrack(s, i + 1);
- path.removeLast();
- }
- }
-
- 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;
- }
- }
比较考字符串操作,回溯比较容易
- class Solution {
- List<String> res = new ArrayList<>();
- public List<String> restoreIpAddresses(String s) {
- StringBuilder sb = new StringBuilder(s);
- backTracking(sb, 0, 0);
- return res;
- }
- void backTracking(StringBuilder s, int start, int dotCount){
- if(dotCount == 3){
- if(isValid(s, start, s.length() - 1)){
- res.add(s.toString());
- }
- return;
- }
- for(int i = start; i < s.length(); i++){
- if(isValid(s, start, i)){
- s.insert(i + 1, '.');
- backTracking(s, i + 2, dotCount + 1);
- s.deleteCharAt(i + 1);
- }else{
- break;
- }
- }
- }
- boolean isValid(StringBuilder s, int start, int end){
- if(start > end) return false;
- if(s.charAt(start) == '0' && start != end) return false;
- int num = 0;
- for(int i = start; i <= end; i++){
- int digit = s.charAt(i) - '0';
- num = num * 10 + digit;
- if(num > 255) return false;
- }
- return true;
- }
- }
可以注释掉 if (start >= nums.length) return;
这行代码,因为、在每一次递归调用中,都会先将当前的 path
加入到结果集中,然后再进行下一层的递归搜索。当 start >= nums.length
时,for 循环条件不满足,递归就自然会结束。
- class Solution {
- List<List<Integer>> res = new ArrayList<>();
- LinkedList<Integer> path = new LinkedList<>();
- public List<List<Integer>> subsets(int[] nums) {
- backtrack(nums,0);
- return res;
- }
- void backtrack(int[] nums, int start) {
- res.add(new ArrayList<>(path));
- //if (start >= nums.length) return;
- for(int i = start; i < nums.length; i++) {
- path.add(nums[i]);
- backtrack(nums, i+1);
- path.removeLast();
- }
- }
- }
和前一天的组合数去重一个套路,理解了树层去重和树枝去重
- class Solution {
- List<List<Integer>> res = new ArrayList<>();
- List<Integer> path = new LinkedList<>();
- boolean[] used;
- public List<List<Integer>> subsetsWithDup(int[] nums) {
- if (nums.length == 0){
- res.add(path);
- return res;
- }
- Arrays.sort(nums);
- used = new boolean[nums.length];
- backtrack(nums, 0);
- return res;
- }
- void backtrack(int[] nums, int start) {
- res.add(new ArrayList<>(path));
- //if (start >= nums.length) return;
- for(int i = start; i < nums.length; i++) {
- if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
- path.add(nums[i]);
- used[i] = true;
- backtrack(nums, i+1);
- path.removeLast();
- used[i] = false;
- }
- }
- }
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