From 97f717d9990c7e7ff2174098f31a2771471e6dfd Mon Sep 17 00:00:00 2001
From: nanhuaibeian <49868746+nanhuaibeian@users.noreply.github.com>
Date: Wed, 12 May 2021 20:56:48 +0800
Subject: [PATCH] =?UTF-8?q?Update=200416.=E5=88=86=E5=89=B2=E7=AD=89?=
 =?UTF-8?q?=E5=92=8C=E5=AD=90=E9=9B=86.md?=
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---
 problems/0416.分割等和子集.md | 34 +++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)

diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md
index 257c4d7d..789a7fb5 100644
--- a/problems/0416.分割等和子集.md
+++ b/problems/0416.分割等和子集.md
@@ -185,7 +185,41 @@ public:
 
 
 Java：
+```Java
+class Solution {
+    public boolean canPartition(int[] nums) {
+        int sum = 0;
+        for (int i : nums) {
+            sum += i;
+        }
+        if ((sum & 1) == 1) {
+            return false;
+        }
+        int length = nums.length;
+        int target = sum >> 1;
+        //dp[j]表示前i个元素可以找到相加等于j情况
+        boolean[] dp = new boolean[target + 1];
+        //对于第一个元素，只有当j=nums[0]时，才恰好填充满
+        if (nums[0] <= target) {
+            dp[nums[0]] = true;
+        }
 
+        for (int i = 1; i < length; i++) {
+            //j由右往左直到nums[i]
+            for (int j = target; j >= nums[i]; j--) {
+                //只有两种情况，要么放，要么不放
+                //取其中的TRUE值
+                dp[j] = dp[j] || dp[j - nums[i]];
+            }
+            //一旦满足，结束，因为只需要找到一组值即可
+            if (dp[target]) {
+                return dp[target];
+            }
+        }
+        return dp[target];
+    }
+}
+```
 
 Python：