[LeetCode] 494. Target Sum 目标和
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.
给一个由非负整数组成的数组，和一个目标值，给数字前面加上正号或负号，然后求和，问和目标值相等的情况有多少。
解法：递归，循环数组里的数字，调用递归函数，分别对目标值进行加上和减去当前数字，再调用递归，这样就会涵盖所有情况，若目标值为0了，则结果res自增1。
Java:
public int findTargetSumWays(int[] nums, int s) {
int sum = 0;
for (int n : nums)
sum += n;
return sum < s || (s + sum) % 2 > 0 ? 0 : subsetSum(nums, (s + sum) >>> 1);
}
public int subsetSum(int[] nums, int s) {
int[] dp = new int[s + 1];
dp[0] = 1;
for (int n : nums)
for (int i = s; i >= n; i--)
dp[i] += dp[i - n];
return dp[s];
}
Python:
class Solution(object):
def findTargetSumWays(self, nums, S):
if not nums:
return 0
dic = {nums[0]: 1, -nums[0]: 1} if nums[0] != 0 else {0: 2}
for i in range(1, len(nums)):
tdic = {}
for d in dic:
tdic[d + nums[i]] = tdic.get(d + nums[i], 0) + dic.get(d, 0)
tdic[d - nums[i]] = tdic.get(d - nums[i], 0) + dic.get(d, 0)
dic = tdic
return dic.get(S, 0)
C++:
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int s) {
int sum = accumulate(nums.begin(), nums.end(), 0);
return sum < s || (s + sum) & 1 ? 0 : subsetSum(nums, (s + sum) >> 1);
}
int subsetSum(vector<int>& nums, int s) {
int dp[s + 1] = { 0 };
dp[0] = 1;
for (int n : nums)
for (int i = s; i >= n; i--)
dp[i] += dp[i - n];
return dp[s];
}
};
C++:
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int S) {
int res = 0;
helper(nums, S, 0, res);
return res;
}
void helper(vector<int>& nums, int S, int start, int& res) {
if (start >= nums.size()) {
if (S == 0) ++res;
return;
}
helper(nums, S - nums[start], start + 1, res);
helper(nums, S + nums[start], start + 1, res);
}
};
C++: 使用dp记录中间值优化
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int S) {
vector<unordered_map<int, int>> dp(nums.size());
return helper(nums, S, 0, dp);
}
int helper(vector<int>& nums, int sum, int start, vector<unordered_map<int, int>>& dp) {
if (start == nums.size()) return sum == 0;
if (dp[start].count(sum)) return dp[start][sum];
int cnt1 = helper(nums, sum - nums[start], start + 1, dp);
int cnt2 = helper(nums, sum + nums[start], start + 1, dp);
return dp[start][sum] = cnt1 + cnt2;
}
};
C++:
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int S) {
int n = nums.size();
vector<unordered_map<int, int>> dp(n + 1);
dp[0][0] = 1;
for (int i = 0; i < n; ++i) {
for (auto &a : dp[i]) {
int sum = a.first, cnt = a.second;
dp[i + 1][sum + nums[i]] += cnt;
dp[i + 1][sum - nums[i]] += cnt;
}
}
return dp[n][S];
}
};
C++:
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int S) {
unordered_map<int, int> dp;
dp[0] = 1;
for (int num : nums) {
unordered_map<int, int> t;
for (auto a : dp) {
int sum = a.first, cnt = a.second;
t[sum + num] += cnt;
t[sum - num] += cnt;
}
dp = t;
}
return dp[S];
}
};
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