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Merge pull request #398 from plin005/master
039 to commit week002 homework (LeetCode_242 & LeetCode_671)
2 parents d51b21e + 7be6014 commit dea8b17

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‎Week_01/id_39/LeetCode_083_039.java‎

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package com.paula.algorithmsAndDataStructure.leetCode_83;
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public class LeetCode_083_039 {
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public ListNode deleteDuplicates(ListNode head) {
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if(head == null) return null;
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ListNode p = head;
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ListNode q = null;
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while(p.next != null) {
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if(p.val == p.next.val) {
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if(q == null || (q != null && q.val != p.val)) {
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q = p;
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}
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}else {
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if(q != null && q.val == p.val) {
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q.next = p.next;
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}
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}
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p = p.next;
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}
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if(q != null && p.val == q.val && p.next == null) {
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q.next = null;
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}
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return head;
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}
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}

‎Week_01/id_39/LeetCode_922_039.java‎

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package com.paula.algorithmsAndDataStructure;
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public class LeetCode_922_039 {
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public int[] sortArrayByParityII(int[] A) {
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int len = A.length;
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int i=0; //偶数下标
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int j=i+1; //奇数下标
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Boolean evenIndexOddValue = false;
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Boolean oddIndexEvenValue = false;
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while (i < len-1 && j < len) {
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evenIndexOddValue = isOdd(A[i]);
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oddIndexEvenValue= isEven(A[j]);
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if(!evenIndexOddValue) i +=2;
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if(!oddIndexEvenValue) j +=2;
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if(evenIndexOddValue && oddIndexEvenValue) {
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int tempVal = A[i];
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A[i] = A[j];
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A[j] = tempVal;
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i += 2;
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j += 2;
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}
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}
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return A;
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}
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public boolean isEven(int value) {
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return value%2 == 0 ? true : false;
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}
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public boolean isOdd(int value) {
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return value%2 == 1 ? true : false;
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}
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}

‎Week_02/id_39/LeetCode_242_039.java‎

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import java.util.HashMap;
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import java.util.Map;
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/**
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*
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* @author Paula
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*
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*
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*/
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class Solution {
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private static void chechString(String s, Map<Character, Integer> smap) {
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for (int i = 0; i < s.length(); i++) {
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if (smap.containsKey(s.charAt(i))) {
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smap.put(s.charAt(i), smap.get(s.charAt(i)) + 1);
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} else {
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smap.put(s.charAt(i), 1);
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}
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}
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}
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public static boolean isAnagram(String s, String t) {
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if (s.equals(t)) {
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return true;
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} else if (null == s || null == t || s.length() != t.length() || s.length() <= 0) {//如果长度不相等，直接返回
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return false;
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}
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Map<Character, Integer> smap = new HashMap<Character, Integer>();
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chechString(s, smap);
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for (int i = 0; i < t.length(); i++) {
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char tmp = t.charAt(i);
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if (smap.containsKey(tmp)) {
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smap.put(tmp, smap.get(tmp) - 1);
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}
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if (0 == smap.get(tmp)) {
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smap.remove(tmp);
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}
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}
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return smap.size() == 0;
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}
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public static void main(String[] args) {
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//("anagram", new HashMap<Character, Integer>());
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System.out.println(isAnagram("anagram", "nagaram"));
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System.out.println(isAnagram("rat", "car"));
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}
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}

‎Week_02/id_39/LeetCode_671_039.java‎

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/**
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*
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* @author Paula
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*
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* 执行结果:
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* 执行用时 : 1 ms, 在Second Minimum Node In a Binary Tree的Java提交中击败了86.02% 的用户
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* 内存消耗 : 34 MB, 在Second Minimum Node In a Binary Tree的Java提交中击败了77.61% 的用户
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*
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* test case:
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* [2,2,5,null,null,5,7]
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* [1,1,3,1,1,3,4,3,1]
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* [5,8,5]
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*/
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class Solution {
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public int findSecondMinimumValue(TreeNode root) {
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if(root == null) return -1;
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//第二小节点值
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int secondMin = -1;
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TreeNode p = root; //第二小的节点
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//检查左节点
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if (p.left != null) {
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//如果左子节点值不等于当前节点值, 则左子节点值必然大于当前节点的值
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if(p.left.val != p.val) secondMin = p.left.val;
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//如果左子节点值不等于当前节点值, 继续遍历左子节点下的分支
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else secondMin = findSecondMinimumValue(p.left);
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}
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//检查右节点
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if (p.right != null) {
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//如果右子节点值不等于当前节点值, 则右子节点值必然大于当前节点的值
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if(p.right.val != p.val) {
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//对比从左子树得到的secondMin和右子节点的值
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if(secondMin == -1 || (secondMin != -1 && p.right.val < secondMin) )
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secondMin = p.right.val;
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//如果右子节点值不等于当前节点值, 继续遍历右子节点下的分支
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} else {
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int tmp = findSecondMinimumValue(p.right);
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if (tmp !=-1 && secondMin !=-1 && tmp < secondMin)
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secondMin = tmp;
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}
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}
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return secondMin;
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}
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}

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