【LeetCode】回溯系列

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17. Letter Combinations of a Phone Number

题目：返回九宫格输入法所有可能的组合（0-9）

思路：回溯——利用递归的方法一层层实现。

public class Solution {    public List<String> letterCombinations(String digits) {        List<String> ret = new LinkedList<>();        if(digits == null || digits.length() == 0) return ret;        String[] index = {" ","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        StringBuilder sb = new StringBuilder();        backtrack(digits, index, sb, 0, ret);        return ret;    }    public void backtrack(String digits, String[] index, StringBuilder sb, int x, List<String> ret){        if(x >= digits.length()){            ret.add(sb.toString());            return;        }        else{            int len = Integer.valueOf(digits.substring(x, x+1));            for(int i = 0; i < index[len].length(); i++){                sb.append(index[len].charAt(i));                backtrack(digits, index, sb, x+1, ret);                sb = sb.deleteCharAt(sb.length()-1);            }            return;        }    }}

39. Combination Sum

题目：找到所有和等于目标值的子集。

思路：回溯

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> ret = new LinkedList<>();        List<Integer> list = new LinkedList<>();        backtrack(ret, list, candidates, target, 0, 0);        return ret;    }    public void backtrack(List<List<Integer>> ret, List<Integer> list, int[] nums, int target, int sum, int n){        if(sum == target){            ret.add(new LinkedList<Integer>(list));            return;        }        else if(sum > target){            return;        }        else{            for(int i = n; i < nums.length; i++){                list.add(nums[i]);                backtrack(ret, list, nums, target, sum+nums[i], i);                list.remove(list.size()-1);            }        }    }}

40. Combination Sum II

题目：数组有重复

思路：在39题基础上排序+去重即可。

public class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        List<List<Integer>> ret = new LinkedList<>();        List<Integer> list = new LinkedList<>();        Arrays.sort(candidates);        backtrack(ret, list, candidates, target, 0, 0);        return ret;    }    public void backtrack(List<List<Integer>> ret, List<Integer> list, int[] nums, int target, int sum, int n){        if(sum == target){            ret.add(new LinkedList<Integer>(list));            return;        }        else if(sum > target){            return;        }        else{            for(int i = n; i < nums.length; i++){                if(i != 0 && i != n && nums[i] == nums[i-1]){                    continue;                }                list.add(nums[i]);                backtrack(ret, list, nums, target, sum+nums[i], i+1);                list.remove(list.size()-1);            }        }    }}

216. Combination Sum III

题目：找到k位之和等于n

思路：感觉跟上面的没啥区别

public class Solution {    public List<List<Integer>> combinationSum3(int k, int n) {        List<List<Integer>> ret = new LinkedList<>();        List<Integer> list = new LinkedList<>();        backtrack(ret, list, k, n, 0, 1);        return ret;    }    public void backtrack(List<List<Integer>> ret, List<Integer> list, int k, int n, int sum, int p){        if(sum == n && list.size() == k){            ret.add(new LinkedList<Integer>(list));            return;        }        else if(sum > n || list.size() > k){            return;        }        else{            for(int i = p; i <= 9; i++){                list.add(i);                backtrack(ret, list, k, n, sum+i, i+1);                list.remove(list.size()-1);            }        }    }}

77. Combinations

题目：找到所有满足固定个数要求的排列。

思路：回溯

public class Solution {    public List<List<Integer>> combine(int n, int k) {        List<List<Integer>> ret = new LinkedList<>();        List<Integer> list = new LinkedList<>();        backtrack(ret, list, n, k, 1);        return ret;    }    public void backtrack(List<List<Integer>> ret, List<Integer> list, int n, int k, int p){        if(list.size() == k){            ret.add(new LinkedList<Integer>(list));            return;        }        else if(list.size() > k){            return;        }        else{            for(int i = p; i <= n; i++){                list.add(i);                backtrack(ret, list, n, k, i+1);                list.remove(list.size()-1);            }        }    }}

78. Subsets

题目：输出所有子集，输入数组不重复

思路：回溯——for循环中设置最远位置

public class Solution {    public List<List<Integer>> subsets(int[] nums) {        List<List<Integer>> ret = new LinkedList<>();        List<Integer> list = new LinkedList<>();        for(int i = 0; i <= nums.length; i++) backtrack(ret, list, nums, 0, i);        return ret;    }    public void backtrack(List<List<Integer>> ret, List<Integer> list, int[] nums, int cur, int n){        if(cur == n){            ret.add(new LinkedList<>(list));            return;        }        else if(cur > n){            return;        }        else{            for(int i = cur; i < n; i++){                list.add(nums[i]);                backtrack(ret, list, nums, i+1, n);                list.remove(list.size()-1);            }        }    }}

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【LeetCode】 回溯系列

【LeetCode】回溯系列