动态规划-377. Combination Sum IV

题目：

Given aninteger arraywith allpositive numbers and no duplicates, find the number of possible combinations that add up to apositive integertarget.

Example:

nums = [1, 2, 3]target = 4The 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.

题目解读：给一个不重复的正整数数组，从该数组中挑选一些数组合相加等于给定的正整数，问这种组合方式有多少种，注意一个trick:数组中的元素是可以反复挑选的。

recursive solution.思路：comb[target] = sum(comb[target - nums[i]]), where 0 <= i < nums.length, and target >= nums[i]这种做法有大量的重复计算。而且出现了Time Limit Exceededclass Solution {    public int combinationSum4(int[] nums, int target) {        if(target == 0) return 1;        int res = 0;        for(int i = 0; i< nums.length; i++){            if(target >= nums[i])                res += combinationSum4(nums,target-nums[i]);        }        return res;    }}

改造成DP,加入备忘录dp[i],累加到i有多少种组合Top-downTime Limit Exceededclass Solution {    public int[] dp;    public int combinationSum4(int[] nums, int target) {        dp = new int[target+1];        //初始化dp        Arrays.fill(dp,-1);        dp[0] = 1;        return helper(nums,target);    }    public int helper(int[] nums,int target){        //先判断是否已经计算过了，避免重复计算        if(dp[target] != -1)             return dp[target];        int res = 0;        for(int i = 0; i < nums.length; i++){            if(target >= nums[i])                res += helper(nums,target-nums[i]);        }        dp[target] = res;        return res;    }}

//bottom_upclass Solution {    public int combinationSum4(int[] nums, int target) {        //定义dp 累加到i有多少种组合        int[] dp = new int[target+1];        dp[0] = 1;        //这种做法没有考虑元素重复        // for(int j = 0; j < nums.length; j++){        //     for(int i = target; i >= nums[j]; i--){        //         dp[i] += dp[i-nums[j]];        //     }        // }        for(int i = 1; i < dp.length; i++){           for(int j = 0; j < nums.length; j++){               if(i >= nums[j])                    dp[i] += dp[i-nums[j]];           }         }        return dp[target];    }}