20191219算法题存档

题目描述

给出两个有序的整数数组A和B，请将数组B合并到数组A中，变成一个有序的数组

注意：

可以假设A数组有足够的空间存放B数组的元素，A和B中初始的元素数目分别为m和n

public class Solution {public void merge(int A[], int m, int B[], int n) {int tempA = --m;int tempB = --n;while(tempA >= 0 && tempB >= 0) {if(A[tempA] >= B[tempB]) {A[tempA + tempB + 1] = A[tempA--];} else {A[tempA + tempB + 1] = B[tempB--];}}if(tempB >= 0) {for(int i = 0; i < tempB + 1; i++) {A[i] = B[i];}}}
}

题目描述

题目给出一个字符串s1，我们可以用递归的方法将字符串分成两个非空的子串来将s1表示成一个二叉树

下面是s1=“great”的一种二叉树的表现形式：

    great↵   /    ↵  gr    eat↵ /     /  ↵g   r  e   at↵           / ↵          a   t

将字符串乱序的方法是：选择任意的非叶子节点，交换它的两个孩子节点。

例如：如果我们选择节点“gr”交换他的两个孩子节点，就会产生一个乱序字符串"rgeat".

    rgeat↵   /    ↵  rg    eat↵ /     /  ↵r   g  e   at↵           / ↵          a   t

我们称"rgeat"是"great"的一个乱序字符串。

类似的：如果我们继续交换“eat”的两个孩子节点和“at”的两个孩子节点，会产生乱序字符串"rgtae".

    rgtae↵   /    ↵  rg    tae↵ /     /  ↵r   g  ta  e↵       / ↵      t   a

我们称"rgtae"是"great"的一个乱序字符串。

给出两个长度相同的字符串s1 和 s2，请判断s2是否是s1的乱序字符串。

public class Solution {public boolean isScramble(String s1, String s2) {if(s1.equals(s2)) {return true;}int[] letters = new int[26];for (int i = 0; i < s1.length(); i++) {letters[s1.charAt(i) - 'a']++;letters[s2.charAt(i) - 'a']--;}for (int i = 0; i < 26; i++) {if (letters[i] != 0) {return false;}}for (int i = 1; i < s1.length(); i++) {if (isScramble(s1.substring(0, i), s2.substring(0, i))&& isScramble(s1.substring(i), s2.substring(i))) {return true;}if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i))&& isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {return true;}}return false;}
}

题目描述

给出一个链表和一个值x，以x为参照将链表划分成两部分，使所有小于x的节点都位于大于或等于x的节点之前。

两个部分之内的节点之间要保持的原始相对顺序。

例如：

给出1->4->3->2->5->2和x = 3,

返回1->2->2->4->3->5.

public class Solution {public ListNode partition(ListNode head, int x) {ListNode rightHead = new ListNode(0);ListNode leftHead = new ListNode(0);ListNode right = rightHead;ListNode left = leftHead;while (head != null) {if (head.val < x) {right.next = head;right = right.next;} else {left.next = head;left = left.next;}head = head.next;}right.next = leftHead.next;left.next = null;return rightHead.next;}
}

题目描述

给出一个只包含0和1的二维矩阵，找出最大的全部元素都是1的长方形区域，返回该区域的面积。

public class Solution {public int maximalRectangle(char[][] matrix) {int result = 0;if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {return result;}int y = matrix.length;int x = matrix[0].length;int[] singleHigh = new int[x];for(int i = 0; i < y; i++) {for(int j = 0; j < x; j++) {if(matrix[i][j] == '1') {singleHigh[j] += 1;} else {singleHigh[j] = 0;}}Stack<Integer> stack = new Stack<>();stack.push(-1);for (int j = 0; j < x; j++) {while (stack.peek() != -1 && singleHigh[j] < singleHigh[stack.peek()]) {int top = stack.pop();result = Math.max(result, (j - 1 - stack.peek()) * singleHigh[top]);}stack.push(j);}while (stack.peek() != -1) {int top = stack.pop();result = Math.max(result, (x - 1 - stack.peek()) * singleHigh[top]);}}return result;}
}

题目描述

给出n个数字，代表直方图的条高，直方图每一条的宽度为1，请计算直方图中最大矩形的面积

public class Solution {public int largestRectangleArea(int[] height) {int result = 0;Stack<Integer> stack = new Stack<>();stack.push(-1);for (int j = 0; j < height.length; j++) {while (stack.peek() != -1 && height[j] < height[stack.peek()]) {int top = stack.pop();result = Math.max(result, (j - 1 - stack.peek()) * height[top]);}stack.push(j);}while (stack.peek() != -1) {int top = stack.pop();result = Math.max(result, (height.length - 1 - stack.peek()) * height[top]);}return result;}
}

题目描述

给出一个排好序的链表，删除链表中的所有重复出现的元素，只保留原链表中只出现一次的元素。

例如：

给出的链表为1->2->3->3->4->4->5, 返回1->2->5.

给出的链表为1->1->1->2->3, 返回2->3.

public class Solution {public ListNode deleteDuplicates(ListNode head) {if(head == null || head.next == null) {return head;}ListNode first = new ListNode(-1);first.next = head;ListNode pre = first;ListNode now = head;while(now != null && now.next != null){if(now.val != now.next.val){pre = now;}else{while(now.next != null && now.val == now.next.val) {now = now.next;}pre.next = now.next;}now = now.next;}return first.next;}}