leetcode39 combination sum

luoyibu 發(fā)布于2019-08-16 10:33 / 3512人閱讀

摘要：在這道題中，我結(jié)合了遞歸的思想來(lái)。就是將當(dāng)前的值作為一個(gè)潛在的結(jié)果值加入一個(gè)結(jié)果數(shù)組將數(shù)組作為當(dāng)前結(jié)果傳入下一輪遞歸。

題目要求

Given a set of candidate numbers (C) (without duplicates) 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.
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]
]

一個(gè)整數(shù)數(shù)組，數(shù)組中的值不重復(fù)，要求在數(shù)組中找到所有的子數(shù)組，子數(shù)組滿足元素的和為目標(biāo)值的條件

思路和代碼

這道題目有一個(gè)標(biāo)簽是backtracking，即在前一種條件的情況下計(jì)算當(dāng)前條件產(chǎn)生的結(jié)果值。在這道題中，我結(jié)合了遞歸的思想來(lái)。就是將當(dāng)前的值作為一個(gè)潛在的結(jié)果值加入一個(gè)結(jié)果數(shù)組將數(shù)組作為當(dāng)前結(jié)果傳入下一輪遞歸。

public class CombinationSum_39 {
    List> result = new ArrayList>();
    public List> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        
        for(int i = 0 ; i temp = new ArrayList();
                temp.add(candidates[i]);
                combinationSum(candidates, i, target-candidates[i], temp);

            }
        }
        return result;
    }
    
    public void combinationSum(int[] candidates, int start, int target, List currentResult){
        for(int i = start ; i < candidates.length ; i++){
            if(candidates[i] == target){
                currentResult.add(candidates[i]);
                result.add(currentResult);
                return;
            }
            if(candidates[i] > target){
                return;
            }
            if(candidates[i] < target){
                List temp = new ArrayList();
                temp.addAll(currentResult);
                temp.add(candidates[i]);
                combinationSum(candidates, i, target-candidates[i], temp);
            }
        }
    }
}

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