动态规划（0-1背包）--- 组合总和

组合总和

377. Combination Sum IV (Medium)

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.

题目描述：

  给定一个值，和一个数组，求数组中元素组合的个数，组合的限制条件是组合中的所有元素和为给定的值。

思路分析：

  动态规划和回溯思想都可以解决。

代码：

DFS

public int res=0;
public int combinationSum4(int []nums,int target){
    if(nums==null||nums.length==0)
        return 0;
    backtracking(nums,0,target);
    return res;
}
public void backtracking(int []nums,int start,int target){
    if(target==0){
        res++;
        return;
    }
    if(target<0)
        return ;
    for(int i=0;i<nums.length;i++){
        backtracking(nums,i,target-nums[i]);
    }
    return;
}

DP

public int combinationSum4(int []nums,int target){
    if(nums==null||nums.length==0)
        return 0;
    int []dp=new int [target+1];
    dp[0]=1;
    for(int i=1;i<=target;i++){
        for(int j=0;j<nums.length;j++){
            if(i>=nums[j])
                dp[i]+=dp[i-nums[j]];
        }
    }
    return dp[target];
}