#feat:leetcode p494题题解

src/leetcode/p494/Solution.java

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+package leetcode.p494;
+
+import java.util.Arrays;
+
+/**
+ * @ProjectName: LeetCode
+ * @FileName: Solution
+ * @Author: 杨逸
+ * @Data:2025/8/12 13:59
+ * @Description: https://leetcode.cn/problems/target-sum/description/
+ * 494. 目标和
+ */
+class Solution {
+    int[] n;
+    int[][] f;
+    public int findTargetSumWays(int[] nums, int target) {
+        /**
+         * 思路：动态规划(记忆化搜索)
+         * p:表示正数的和
+         * s:表示nums的和
+         * s-p:表示负数的和
+         *所以:p - (s - p) = target
+         * p = (target + s) / 2
+         * 又因为nums是整数,所以(target + s) / 2必须是正整数
+         * 问题转化为从nums中选出一些数,使其和为(target + s) / 2的方案数
+         */
+        n  = nums;
+        target += Arrays.stream(nums).sum();
+        if (target % 2 != 0 || target < 0) {
+            return 0;
+        }
+        target /= 2;
+        f = new int[nums.length][target+1];
+        for (int i = 0; i < f.length; i++) {
+            Arrays.fill(f[i],-1);
+        }
+        return dfs(nums.length-1,target);
+    }
+
+    private int dfs(int i, int target) {
+        //边界条件
+        if(i<0){
+            //选出的数刚好等于p,合法方案数+1
+            return target == 0 ? 1 : 0;
+        }
+        //之前计算过
+        if(f[i][target] != -1){
+            return f[i][target];
+        }
+        //不选的方案数计算
+        int res =  dfs(i-1,target);
+        //能选方案数
+        if(target >= n[i]){
+            res += dfs(i-1,target-n[i]);
+        }
+        //记忆化
+        return f[i][target] = res;
+    }
+}

src/leetcode/p494/SolutionTest.java

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+package leetcode.p494;
+
+import org.junit.Test;
+
+/**
+ * @ProjectName: LeetCode
+ * @FileName: SolutionTest
+ * @Author: 杨逸
+ * @Data:2025/8/12 14:43
+ * @Description: 
+ */
+public class SolutionTest {
+
+    @Test
+    public void findTargetSumWays() {
+        int target = 3;
+        int[] nums = {1, 1, 1, 1, 1};
+        int targetSumWays = new Solution().findTargetSumWays(nums, target);
+        System.out.println(targetSumWays);
+    }
+}