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Dan LI
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_posts/2020-11-13-Naive Dynamic Programming.md

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@@ -12,7 +12,7 @@ num_coins = min(1+num_coins(original_change-1), 1+num_coins(original_change-5),
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## 2. 问题的递归解决
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从公式可以看到问题的递归特性,脑海中也可以想象出大致的时间复杂度。如下图所示:
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![call tree](./assets/callTree.png)
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![call tree](https://github.com/ldmax/ldmax.github.io/blob/master/_posts/assets/callTree.png)
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树的每个节点理论上会有1~4个子节点,最大层数为n1 = original_change / min(coin_list), 最小层数n2 = original_change / max(coin_list)。original_change是找零金额,coin_list是可用的硬币金额列表。
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