可以使用方法：回溯法、DP。

問題描述：

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.

窮舉法，判斷每一種組合是否滿足條件。很容易想象有O（2^n）種組合方式，對於每種組合形式，如果：1、獨立的累加，那麼演算法的最終時間複雜度為O(n*2^n)，對應三層for迴圈；2、用一變數記錄，則，最終複雜度為O(2^n),對應兩層for迴圈。

在這使用dfs，時間複雜度為O(2^n),但包含一些剪枝。實際用時310ms。C程式碼：

void dfs(int* nums, int numsSize, int Sum, int begin, int S, int* ret){ //Sum:儲存上個狀態的和 begin:下次迴圈的起始index
    for(int i = begin; i < numsSize; ++i){
        int tmp = Sum - 2*nums[i];                         
        if(tmp == S) (*ret)++; 
        if(tmp >= S) dfs(nums, numsSize, tmp, i+1, S, ret);//因為題目中說的是non-negative，包括0，所以得>=而不是>。
    }                                                      //考慮{ [0,0,0,0,0,0,0,0,1] 1 }例子。
}
int findTargetSumWays(int* nums, int numsSize, int S) {
    int ret = 0, Sum = 0;
    for(int i = 0; i < numsSize; ++i) Sum += nums[i];
    if(Sum == S) ret++;
    dfs(nums, numsSize, Sum, 0, S, &ret);
    return ret;
}
 

sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
                       2 * sum(P) = target + sum(nums)
所以原問題就轉化成了子陣列和問題（類似的：最大連續子陣列和，和為S的連續正數序列）-----找到一個子集P使得sum(P) = (target + sum(nums)) / 2。注意，target + sum(nums)必須為偶數。

int subsetSum(int* nums, int numsSize, int target) {
    int ret;
    int* dp = (int*)malloc((target+1)*sizeof(int));
    memset(dp, 0, (target+1)*sizeof(int));
    dp[0] = 1;
    for(int i = 0; i < numsSize; ++i)
        for(int j = target; j >= nums[i]; --j)
            dp[j] += dp[j-nums[i]];
    ret = dp[target];
    free(dp);
    return ret;
}

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