From 3223112124c71c21fef1354a451b12a64c118f7d Mon Sep 17 00:00:00 2001
From: KiloG <33695125+JOJO0527@users.noreply.github.com>
Date: Fri, 7 Oct 2022 21:26:39 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0Go=E7=9A=84=E5=8F=A6=E5=A4=96?=
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可以将题目转化为“求装满背包s的前几位字符的方式有几种”， 然后判断最后dp[len(s)]是否大于0就可。
---
 problems/0139.单词拆分.md | 15 +++++++++++++++
 1 file changed, 15 insertions(+)

diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md
index d60fd46c..5226502a 100644
--- a/problems/0139.单词拆分.md
+++ b/problems/0139.单词拆分.md
@@ -344,6 +344,21 @@ func wordBreak(s string,wordDict []string) bool  {
 	}
 	return dp[len(s)]
 }
+// 转化为 求装满背包s的前几位字符的方式有几种
+func wordBreak(s string, wordDict []string) bool {
+	// 装满背包s的前几位字符的方式有几种
+	dp := make([]int, len(s)+1)
+	dp[0] = 1
+	for i := 0; i <= len(s); i++ { // 背包
+		for j := 0; j < len(wordDict); j++ { // 物品
+			if i >= len(wordDict[j]) && wordDict[j] == s[i-len(wordDict[j]):i] {
+				dp[i] += dp[i-len(wordDict[j])]
+			}
+		}
+	}
+
+	return dp[len(s)] > 0
+}
 ```
 
 Javascript：