494.Target Sum

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.

 

sum(P)-sum(n)=target;

sum(P)-sum(n)+sum(P)+sum(n)=target+sum(P)+sum(n);

2*sum(P)=target+sum(nums);

sum(P)=(target+sum(nums))/2;

 

最后转换为求sum(P)，即数组中有几个组合加起来等于sum(P)

dp[i]的意思是：元素和为i的话有几种情况，

当前元素为4时，则dp[i] = dp[i] + dp[i-4];

dp[9] = dp[9] + dp[9-4];

dp[8] = dp[8] + dp[8-4];

dp[7] = dp[7] + dp[7-4];

dp[6] = dp[6] + dp[6-4];

dp[5] = dp[5] + dp[5-4];

dp[4] = dp[4] + dp[4-4];

 

和题目416解法一样

 

public static int findTargetSumWays(int[] nums, int target) {
  int sum = 0;
  for (int i = 0; i < nums.length; i++) {
    sum += nums[i];
  }
  if (target > sum || (sum + target) % 2 == 1)
    return 0;
  return subsetSum(nums, (sum + target) / 2);
}

private static int subsetSum(int[] nums, int target) {
  int[] dp = new int[target + 1];
  dp[0] = 1; //C(0,0)=1
  for (int i = 0; i < nums.length; i++) {
    for (int j = target; j >= nums[i]; j--) { // 需要target从大到小遍历，如果从小到大遍历，则每一次内层循环，都会给dp[target]增加值
        dp[j] += dp[j - nums[i]];
    }
  }
  return dp[target];
}