• [LeetCode#39]Combination Sum


    Problem:

    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:

    • All numbers (including target) will be positive integers.
    • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
    • The solution set must not contain duplicate combinations.

    For example, given candidate set 2,3,6,7 and target 7
    A solution set is: 
    [7] 
    [2, 2, 3]

    Analysis:

    This problem is a typical DFS problem, it is simple, but it is very very useful. It actually shares same skills we have used in two sum problem: 
    1. How to take advantage of a sorted array to aovid duplicates?
    2. How to guarantee the results in no-descending order?
    
    They all could solve through the skill we have used in three sum problem: sort the array first. But there is also an obvious difference between this problem with two sum problem. 
    -------------------------------------------------------------------------------------------------------------------------
    3 sum problem: there could be duplicates in the array, we should use some skills to avoid the elements with value appear was used on the same index many times. 
    -------------------------------------------------------------------------------------------------------------------------
    combinaionSum problem: no duplicates in the candidates set, and we allow to use each element many times. (This free us from checking iff nums[i] == nums[i-1]). 
    -------------------------------------------------------------------------------------------------------------------------
    
    Basic idea:
    step 1: sort the candidates in no-descending order.
    Arrays.sort(candidates);
    step 2: get a number from the sorted array, then continue the search along the elements after it(inclusive, since each element could be used multi times)
    Note: the "inclusive" would allow the same element to be used at other indexes, but not repeat on the same index. 
    A tree structure is good start for understanding!
    for (int i = start; i < candidates.length; i++) {
        item.add(candidates[i]);
        searchPath(candidates, i, target-candidates[i], item, ret);
        item.remove(item.size()-1);
    }
    
    Good programming habit:
    Don't do the complex checking when forking!!! Stop the wrong froking branches at base case!!!
    if (target == 0)
        ret.add(new ArrayList<Integer> (item));
    if (target < 0)
        return;
        
        
    Two repeatives in such problem:
    1. the element with the same value were used on the same index many times.
        1.1 iff there are duplicates in array (nums[i] != num[i-1])
        1.2 iff there are not duplicates in array (easy, but should allow it at next index)
        
    2. mix the order of the same result.
    [2 3] [3 2]
    The problem was caused by following logic:
    for (int i = 0; i < candidates.length; i++) {
        ...
        searchPath(candidates, i, target-candidates[i], item, ret);
    }
    It must cause duplicates!
    
    
    The great way to solve above prolem is using sorting candidates array!
    And use the fix:
    for (int i = start; i < candidates.length; i++) {
        ...
    }
    start from "start", rather than the elements before start.

    Solution:

    public class Solution {
        public List<List<Integer>> combinationSum(int[] candidates, int target) {
            if (candidates == null)
                throw new IllegalArgumentException("candidates is null");
            List<List<Integer>> ret= new ArrayList<List<Integer>> ();
            if (candidates.length == 0 || target < 0)
                return ret;
            ArrayList<Integer> item = new ArrayList<Integer> ();
            Arrays.sort(candidates);
            searchPath(candidates, 0, target, item, ret);
            return ret;
        }
        
        private void searchPath(int[] candidates, int start, int target, List<Integer> item, List<List<Integer>> ret) {
            if (target == 0)
                ret.add(new ArrayList<Integer> (item));
            if (target < 0)
                return;
            for (int i = start; i < candidates.length; i++) {
                item.add(candidates[i]);
                searchPath(candidates, i, target-candidates[i], item, ret);
                item.remove(item.size()-1);
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/airwindow/p/4788592.html
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