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Author: assassint2017

Target Sum/Target_Sum.cpp

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+// 第一种思路，暴力回溯，果然超时了
+class Solution 
+{
+public:
+    int findTargetSumWays(vector<int>& nums, int target) 
+    {
+        return dfs(nums, target, 0);
+    }
+private:
+    int dfs(const vector<int>& nums, int target, int start)
+    {
+        if (start == nums.size() - 1)
+        {
+            int res = 0;
+            
+            target == nums.back() ? ++res : res;
+            target == -nums.back() ? ++res : res;
+            
+            return res;
+        }
+        else
+            return dfs(nums, target - nums[start], start + 1) + dfs(nums, target + nums[start], start + 1);
+    }
+};
+
+// 第二种思路，使用动态规划
+// Runtime: 44 ms, faster than 63.27% of C++ online submissions for Target Sum.
+// Memory Usage: 21.1 MB, less than 30.77% of C++ online submissions for Target Sum.
+
+class Solution 
+{
+public:
+    int findTargetSumWays(vector<int>& nums, int S) 
+    {
+        int sum = 0;
+        for (auto item : nums)
+            sum += item;
+        
+        int maxValue = sum, minValue = -sum;
+        
+        // 边界条件处理
+        if (nums.size() == 0 || S < minValue || S > maxValue)
+            return 0;
+        
+        vector<vector<int>> dp(nums.size(), vector<int>(maxValue - minValue + 1, 0));
+        
+        for (int i = 0; i < dp.size(); ++i)
+        {
+            for (int j = 0; j < dp[0].size(); ++j)
+            {
+                if (i == 0)
+                {
+                    int temp = 0;
+                    nums.back() == j + minValue ? ++temp : 0;
+                    -nums.back() == j + minValue ? ++temp : 0;
+                    dp[i][j] = temp;
+                }
+                else
+                {
+                    int index = nums.size() - i - 1;
+                    if (j - nums[index] >= 0 && j - nums[index] < dp[0].size())
+                        dp[i][j] += dp[i - 1][j - nums[index]];
+                    if (j + nums[index] >= 0 && j + nums[index] < dp[0].size())
+                        dp[i][j] += dp[i - 1][j + nums[index]];
+                }
+            }
+        }
+        return dp[dp.size() - 1][S - minValue];
+    }
+};