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Commitf29db46

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[LEET-504] add 504
1 parent9c3532f commitf29db46

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‎leetcode-algorithms/README.md

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|506|[Relative Ranks](https://leetcode.com/problems/relative-ranks/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/RelativeRanks.java)| O(nlogn)|O(n)| Easy|
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|505|[The Maze II](https://leetcode.com/problems/the-maze-ii/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TheMazeII.java) | O(m*n) |O(m*n) | Medium| BFS
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|504|[Base 7](https://leetcode.com/problems/base-7/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/Base7.java)| O(1)|O(1)| Easy|
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|503|[Next Greater Element II](https://leetcode.com/problems/next-greater-element-ii/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/NextGreaterElementII.java) | O(n) |O(n) | Medium| Stack
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|501|[Find Mode in Binary Tree](https://leetcode.com/problems/find-mode-in-binary-tree/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/FindModeinBinaryTree.java) | O(n) |O(k) | Easy| Binary Tree
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|500|[Keyboard Row](https://leetcode.com/problems/keyboard-row/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/KeyboardRow.java)| O(n)|O(1)| Easy|
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|499|[The Maze III](https://leetcode.com/problems/the-maze-iii/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/TheMazeIII.java) | O(m*n) |O(m*n) | Hard| BFS
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packagecom.stevesun.solutions;
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importjava.util.Stack;
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/**
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* Given a circular array (the next element of the last element is the first element of the array),
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* print the Next Greater Number for every element.
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* The Next Greater Number of a number x is the first greater number to its traversing-order next in the array,
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* which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.
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Example 1:
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Input: [1,2,1]
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Output: [2,-1,2]
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Explanation: The first 1's next greater number is 2;
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The number 2 can't find next greater number;
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The second 1's next greater number needs to search circularly, which is also 2.
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Note: The length of given array won't exceed 10000.
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*/
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publicclassNextGreaterElementII {
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//Credit: https://discuss.leetcode.com/topic/77881/typical-ways-to-solve-circular-array-problems-java-solution
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//Note: we store INDEX into the stack, reversely, the larger index put at the bottom of the stack, the smaller index at the top
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publicint[]nextGreaterElements(int[]nums) {
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if (nums ==null ||nums.length ==0)returnnums;
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intlen =nums.length;
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Stack<Integer>stack =newStack<>();
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for (inti =len-1;i >=0;i--) {
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stack.push(i);//push all indexes into the stack reversely
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}
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int[]result =newint[len];
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for (inti =len-1;i >=0;i--) {
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result[i] = -1;//initialize it to be -1 in case we cannot find its next greater element in the array
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while (!stack.isEmpty() && (nums[stack.peek()] <=nums[i])) {
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stack.pop();
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}
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if (!stack.isEmpty()) {
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result[i] =nums[stack.peek()];
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}
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stack.push(i);
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}
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returnresult;
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}
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//credit: https://leetcode.com/articles/next-greater-element-ii/
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publicint[]nextGreaterElements_editorial_solution(int[]nums) {
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int[]result =newint[nums.length];
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Stack<Integer>stack =newStack<>();
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for (inti =nums.length*2-1;i>=0 ;i--) {
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while (!stack.isEmpty() &&nums[stack.peek()] <=nums[i%nums.length]) {
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stack.pop();
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}
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result[i%nums.length] =stack.isEmpty() ? -1 :nums[stack.peek()];
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stack.push(i%nums.length);
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}
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returnresult;
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}
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}
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packagecom.stevesun;
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importcom.stevesun.solutions.NextGreaterElementII;
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importorg.junit.Before;
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importorg.junit.BeforeClass;
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importorg.junit.Test;
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importstaticorg.junit.Assert.assertArrayEquals;
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publicclassNextGreaterElementIITest {
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privatestaticNextGreaterElementIItest;
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privatestaticint[]nums;
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privatestaticint[]expected;
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privatestaticint[]actual;
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@BeforeClass
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publicstaticvoidsetup(){
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test =newNextGreaterElementII();
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}
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@Before
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publicvoidsetupForEachTest(){
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expected =newint[]{};
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nums =newint[]{};
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}
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@Test
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publicvoidtest1(){
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nums =newint[]{1,2,1};
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expected =newint[]{2, -1,2};
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actual =test.nextGreaterElements(nums);
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assertArrayEquals(expected,actual);
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}
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}

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