7th Feb: 刷题笔记

1/ Maximum Depth of Binary TreeGiven a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int maxDepth(TreeNode root) {        if (root == null) {            return 0;        }                int left = maxDepth(root.left);        int right = maxDepth(root.right);        int ans = 1 + Math.max(left, right);                return ans;    }}
2/ Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.For example,Given [3,2,1,5,6,4] and k = 2, return 5.Note: You may assume k is always valid, 1 ≤ k ≤ array's length.
最简单的做法：
public class Solution {    public int findKthLargest(int[] nums, int k) {        Arrays.sort(nums);        return nums[nums.length - k];    }}
或者利用PRiority Queue实现大顶堆，接着去到第k个。
快速排序的做法（明天补充）：
3/ Minstack （见5th Feb的笔记）
4/  Implement strStr()Implement strStr().Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
public class Solution {    public int strStr(String haystack, String needle) {        if (haystack == null || needle == null) {            return -1;        }                for (int i = 0; i < haystack.length() - needle.length() + 1; i++) {            int j = 0;            for (j = 0; j < needle.length(); j++) {                if (haystack.charAt(i + j) != needle.charAt(j)) {                    break;                }            }            if (j == needle.length()) {//错误                    return i;            }        }        return -1;    }}这道题，之前把if写在for循环里的话，就错了。
5/ Rectangle Area
Find the total area covered by two rectilinear rectangles in a 2D plane.
Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.Assume that the total area is never beyond the maximum possible value of int.
public class Solution {    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {                int areaOfSqrA = (C-A) * (D-B);         int areaOfSqrB = (G-E) * (H-F);                int left = Math.max(A, E);        int right = Math.min(G, C);        int bottom = Math.max(F, B);        int top = Math.min(D, H);                //If overlap        int overlap = 0;        if(right > left && top > bottom)             overlap = (right - left) * (top - bottom);                return areaOfSqrA + areaOfSqrB - overlap;    }}6/ Rotate ArrayRotate an array of n elements to the right by k steps.For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
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10/ Given a map of task and its corresponding cooldown period(unordered_map<char, int> tasks), and a sequence of task(such as "ABCDABC"), all tasks are the unit running time and the machine can run a task one time. Return the total time needed to finish all the sequence tasks.
类似的LeetCode题目（Hard 难度，放弃）：https://leetcode.com/problems/rearrange-string-k-distance-apart/
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