From 02fe7d6e79d8c9c644e842dbc12a113bbfce0809 Mon Sep 17 00:00:00 2001 From: qyg <1600314850@qq.com> Date: Mon, 27 Dec 2021 10:07:19 +0800 Subject: [PATCH] =?UTF-8?q?0392.=E5=88=A4=E6=96=AD=E5=AD=90=E5=BA=8F?= =?UTF-8?q?=E5=88=97=EF=BC=9A=E8=B0=83=E6=95=B4=E7=AC=94=E8=AF=AF=EF=BC=8C?= =?UTF-8?q?=E4=BC=98=E5=8C=96=E6=8E=92=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0392.判断子序列.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0392.判断子序列.md b/problems/0392.判断子序列.md index ecde2448..d64e1fd0 100644 --- a/problems/0392.判断子序列.md +++ b/problems/0392.判断子序列.md @@ -81,13 +81,13 @@ if (s[i - 1] != t[j - 1]),此时相当于t要删除元素,t如果把当前 **其实这里只初始化dp[i][0]就够了,但一起初始化也方便,所以就一起操作了**,代码如下: -``` +```CPP vector> dp(s.size() + 1, vector(t.size() + 1, 0)); ``` 4. 确定遍历顺序 -同理从从递推公式可以看出dp[i][j]都是依赖于dp[i - 1][j - 1] 和 dp[i][j - 1],那么遍历顺序也应该是从上到下,从左到右 +同理从递推公式可以看出dp[i][j]都是依赖于dp[i - 1][j - 1] 和 dp[i][j - 1],那么遍历顺序也应该是从上到下,从左到右 如图所示: