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README.md

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@@ -58,6 +58,7 @@ _If you like this project, please leave me a star._ ★
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|1055|[Fixed Point](https://leetcode.com/problems/fixed-point/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1055.java) | |Easy||
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|1051|[Height Checker](https://leetcode.com/problems/height-checker/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1051.java) | |Easy||
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|1047|[Remove All Adjacent Duplicates In String](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1047.java) | |Easy||
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|1046|[Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1046.java) | |Easy||
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|1043|[Partition Array for Maximum Sum](https://leetcode.com/problems/partition-array-for-maximum-sum/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1043.java) | |Medium|DP|
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|1038|[Binary Search Tree to Greater Sum Tree](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1038.java) | |Medium|DFS, tree|
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|1037|[Valid Boomerang](https://leetcode.com/problems/valid-boomerang/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1037.java) | |Easy|Math|
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package com.fishercoder.solutions;
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import java.util.PriorityQueue;
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/**
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* 1046. Last Stone Weight
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*
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* We have a collection of rocks, each rock has a positive integer weight.
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*
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* Each turn, we choose the two heaviest rocks and smash them together.
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* Suppose the stones have weights x and y with x <= y. The result of this smash is:
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*
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* If x == y, both stones are totally destroyed;
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* If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
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* At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)
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*
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* Example 1:
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* Input: [2,7,4,1,8,1]
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* Output: 1
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* Explanation:
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* We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
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* we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
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* we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
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* we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.
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*
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* Note:
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* 1 <= stones.length <= 30
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* 1 <= stones[i] <= 1000
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* */
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public class _1046 {
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public static class Solution1 {
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public int lastStoneWeight(int[] stones) {
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PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
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for (int stone : stones) {
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heap.offer(stone);
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}
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while (!heap.isEmpty()) {
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if (heap.size() >= 2) {
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int one = heap.poll();
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int two = heap.poll();
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int diff = one - two;
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heap.offer(diff);
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} else {
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return heap.poll();
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}
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}
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return -1;
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}
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}
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}
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package com.fishercoder;
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import com.fishercoder.solutions._1046;
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import org.junit.BeforeClass;
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import org.junit.Test;
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import static junit.framework.Assert.assertEquals;
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public class _1046Test {
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private static _1046.Solution1 solution1;
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@BeforeClass
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public static void setup() {
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solution1 = new _1046.Solution1();
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}
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@Test
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public void test1() {
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assertEquals(1, solution1.lastStoneWeight(new int[]{2, 7, 4, 1, 8, 1}));
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}
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}

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