LeetCode: Target Sum

最新推荐文章于 2022-06-10 11:12:05 发布

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最新推荐文章于 2022-06-10 11:12:05 发布

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分类专栏： LeetCode 文章标签： LeetCode c语言 target sum dynamic programming

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探讨了如何通过选择加减号来使整数列表的总和等于目标值的问题，并将其转换为子集求和问题，提供了具体的算法实现。

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You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 

-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

Note:

The length of the given array is positive and will not exceed 20.
The sum of elements in the given array will not exceed 1000.
Your output answer is guaranteed to be fitted in a 32-bit integer.

Analysis:

The recursive solution is very slow, because its runtime is exponential

The original problem statement is equivalent to:
Find a subset of nums that need to be positive, and the rest of them negative, such that the sum is equal to target

Let P be the positive subset and N be the negative subset
For example:
Given nums = [1, 2, 3, 4, 5] and target = 3 then one possible solution is +1-2+3-4+5 = 3
Here positive subset is P = [1, 3, 5] and negative subset is N = [2, 4]

Then let's see how this can be converted to a subset sum problem:

                  sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
                       2 * sum(P) = target + sum(nums)

So the original problem has been converted to a subset sum problem as follows:
Find a subset P of nums such that sum(P) = (target + sum(nums)) / 2

Note that the above formula has proved that target + sum(nums) must be even

int subsetSum(int*nums, int numsSize, int s) {
    int* dp = (int*)malloc((s + 1) * sizeof(int));
    for (int i = 0; i < s + 1; ++i) {
        dp[i] = 0;
    }
    // for (int i = 0; i < numsSize; ++i) {
    //     dp[nums[i]] = 1;
    // }
    dp[0] = 1;
    
    for (int n = 0; n < numsSize; ++n) {
        for (int i = s; i >= nums[n]; --i) {
            dp[i] += dp[i - nums[n]];
        }
    }
    
    return dp[s];
}

int findTargetSumWays(int* nums, int numsSize, int S) {
    
    int sum = 0;
    for (int i = 0; i < numsSize; ++i) {
        sum += nums[i];
    }
    
    if (sum < S || (sum + S) % 2 == 1) {
        return 0;
    }
    
    
    
    return subsetSum(nums, numsSize, (sum + S) / 2);
    
}