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leetCode_Solution_Record
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9812fab4
authored
Oct 04, 2020
by
李楚霏
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学习总结
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### 第6周学习总结
#### 动态规划
-
关键点
-
分解问题为子问题;
-
存储中间状态;
-
递推公式;
-
用一个二维路径来表示(不一定)
#### 题型总结
-
唯一路径
-
cpp组织,要注意二维数组的初始化;
-
确定左边界和上边界,dp数组的第一行和第一列的元素特殊处理,其他元素则按照递推公式来获得
-
唯一路径ii
-
唯一路径的扩展的基础上加了一个判断当前是否碰到障碍的条件;
-
注意一下js 初始化二维数组的写法
-
解码方法
-
-
求最小路径之和
-
这道题也是有左边界和上边界,第一行和第一列的元素由于只能从当前行
\列
的上一个元素获得,所以它们的递推公式为 dp
[
i,0
]
=dp
[
i-1, 0
]
+此坐标下的路径权重
\
dp
[
0,j
]
=dp
[
0, j-1
]
+此坐标下的路径权重
-
其他元素则就是从上或左的元素的最小值+此坐标下的路径权重,即dp
[
i, j
]
= min(dp
[
i-1,j
]
, dp
[
i, j-1
]
) + 此坐标下的路径权重
\ No newline at end of file
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