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Combination Sum
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[ [7], [2, 2, 3] ]
用for循环+ 递归的方式求解
for 循环套在外层，表示遍历数组的第i个数字
内层递归表示结果list里的第i个数字（比如第一层递归是寻找list里的第一个数字）
1 public class Solution { 2 public List<List<Integer>> combinationSum(int[] candidates, int target) { 3 int size = candidates.length; 4 List<List<Integer>> res = new ArrayList<List<Integer>>(); 5 List<Integer> list = new ArrayList<Integer>(); 6 helper (res, list, candidates, target, 0); 7 return res; 8 } 9 private static void helper(List<List<Integer>> res,List<Integer> list, int[] candidates, int target, int pos) { 10 if (target == 0) { 11 res.add(new ArrayList<Integer>(list)); 12 return; 13 14 } 15 16 if (target < 0) { 17 return; 18 } 19 int prev = -1; 20 for (int i = pos; i < candidates.length; i++) { 21 if(prev == candidates[i]) { 22 continue; 23 } 24 list.add(candidates[i]); 25 helper(res, list, candidates, target - candidates[i], i); 26 prev = candidates[i]; 27 list.remove(list.size() - 1); 28 } 29 } 30 }
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Combination Sum 2
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
这个题目要注意下面的错误情况
用递归寻找第一个元素时，如果之前的结果里已经把1当作第一个元素，然后pop出去了，那么下次的1再想做为第一个元素的话就直接continue，这样就避免了2个[1,2,5]的情况，也就是说，在for循环里面不允许有相同的元素！只有第一次出现才做为valid
1 public class Solution { 2 public List<List<Integer>> combinationSum2(int[] candidates, int target) { 3 int size = candidates.length; 4 Arrays.sort(candidates); 5 List<List<Integer>> res = new ArrayList<List<Integer>>(); 6 List<Integer> list = new ArrayList<Integer>(); 7 helper (res, list, candidates, target, 0); 8 return res; 9 } 10 private static void helper(List<List<Integer>> res,List<Integer> list, int[] candidates, int target, int pos) { 11 if (target == 0) { 12 res.add(new ArrayList<Integer>(list)); 13 return; 14 15 } 16 if (target < 0) { 17 return; 18 } 19 int prev = -1; 20 for (int i = pos; i < candidates.length; i++) { 21 if (candidates[i] > target) { 22 break; 23 } 24 25 if (prev != -1 && prev == candidates[i]) { 26 continue; 27 } 28 29 if (candidates[i] != -1) { 30 list.add(candidates[i]); 31 helper(res, list, candidates, target - candidates[i], i + 1); 32 prev = candidates[i]; 33 list.remove(list.size() - 1); 34 } 35 } 36 } 37 }
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Combination Sum 3
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
1 public class Solution { 2 public List<List<Integer>> combinationSum3(int k, int n) { 3 List<List<Integer>> res = new ArrayList<List<Integer>>(); 4 List<Integer> list = new ArrayList<Integer>(); 5 helper (res, list, k, n, 1); 6 return res; 7 } 8 private static void helper(List<List<Integer>> res,List<Integer> list, int k, int n, int current) { 9 if (n == 0 && list.size() == k) { 10 res.add(new ArrayList<Integer>(list)); 11 return; 12 13 } 14 if (list.size() == k || n < 0) { 15 return; 16 } 17 18 19 for (int i = current; i < 10; i++) { 20 list.add(i); 21 helper(res, list, k, n - i, i + 1); 22 list.remove(list.size() - 1); 23 } 24 } 25 }
相当于[1,2,3,4,5,6,7,8,9] 的数组，然后增加了k的控制条件
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Combination Sum IV
Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.
Example:
nums = [1, 2, 3] target = 4 The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations. Therefore the output is 7.
Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?
巨坑，只需要返回结果个数，那么其实是个动态规划的题目，类似于爬楼梯！！！
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原文地址：http://www.cnblogs.com/jiangchen/p/5930049.html
