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README.md

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|199|[Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_199.java)| O(n)|O(n)| Medium | BFS
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|198|[House Robber](https://leetcode.com/problems/house-robber/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_198.java)| O(n)|O(n)| Easy | DP
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|191|[Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_191.java)| O(n)|O(1)| Easy | Bit Manipulation
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|190|[Reverse Bits](https://leetcode.com/problems/reverse-bits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/ReverseBits.java)| O(n)|O(1)| Easy | Bit Manipulation
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|190|[Reverse Bits](https://leetcode.com/problems/reverse-bits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_190.java)| O(n)|O(1)| Easy | Bit Manipulation
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|189|[Rotate Array](https://leetcode.com/problems/rotate-array/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_189.java)| O(n)|O(n), could be optimized to O(1) | Easy
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|188|[Best Time to Buy and Sell Stock IV](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_188.java)| O(n*k)|O(n*k) | Hard | DP
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|187|[Repeated DNA Sequences](https://leetcode.com/problems/repeated-dna-sequences/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_187.java)| O(n)|O(n) | Medium

src/main/java/com/fishercoder/solutions/ReverseBits.java renamed to src/main/java/com/fishercoder/solutions/_190.java

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Follow up:
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If this function is called many times, how would you optimize it?*/
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public class ReverseBits {
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public class _190 {
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/**This post: http://stackoverflow.com/questions/2811319/difference-between-and
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* gives a good explanation between logical right shift: ">>>" and arithmetic right shift: ">>".
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* Basically, ">>" preserves the most left bit and treats it as the sign for this number,
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public static void main(String...strings){
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System.out.println(Integer.toBinaryString(4));
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ReverseBits test = new ReverseBits();
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_190 test = new _190();
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int n = 1;
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System.out.println(test.reverseBits(n));
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System.out.println(Integer.parseInt("11000", 2));

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