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/** 题意：求出给定序列的最长递增子序列的长度，给定序列不是有序的，子序列不是子数组，元素在原数组中不必是连续的**//** solutions1: 时间复杂度O(n^2),空间复杂度O(n)* 新建一个辅助数组h,h[i]表示以nums[i]结尾的最长递增子序列的长度* h[i]值的确定，需要nums[i]与nums[j](0<=j= num) {int pos = getPos(h, left, mid – 1, num);if (pos == -1)return mid;return pos;} else {return getPos(h, mid + 1, right, num);}}return -1;}public static void main(String args){LongestSubSequence l = new LongestSubSequence;int nums = {1,3,6,7,9,4,10,5,6};l.getLonggestLen(nums);}}View Code回归到本题的解法。

/** @lc app=leetcode.cn id=287 lang=java** [287] 寻找重复数*/class Solution {public int findDuplicate(int nums) {Arrays.sort(nums);int n = 0;for (int i = 1; i < nums.length; i++) {n = nums[i - 1] ^ nums[i];if (n == 0) {return nums[i];}}return nums[0];}}找到所有数组中消失的数字把值当做索引，把对应索引的值变化成负数，最后没变化的值,就是缺少对应索引的值。

“`C #include double My_variable = 3.0;int fact(int n) {     if (n <= 1) return 1;     else return n*fact(n-1);}int my_mod(int x, int y) {     return (x%y);}char *get_time {     time_t ltime;     time(&ltime);     return ctime(&ltime);} 编译它。

static final class FairSync extends Sync {protected final boolean tryAcquire(int acquires) {final Thread current = Thread.currentThread;int c = getState;if (c == 0) {// 关键差别就是是否检测hasQueuedPredecessorsif (!hasQueuedPredecessors &&compareAndSetState(0, acquires)) {setExclusiveOwnerThread(current);return true;}}else if (current == getExclusiveOwnerThread) {int nextc = c + acquires;if (nextc < 0)throw new Error("Maximum lock count exceeded");setState(nextc);return true;}return false;}}大多数的同步器都存在相同的内在逻辑和实现模式：

(!equal0(a) && equal24(b/a)));}for(int i = 0; i tmp;for(int k = 0; k & nums) {vector t(nums.begin, nums.end);return helper(t);}};附上输出所有可能的解的程序：

import java.util.Arrays;import java.util.Comparator;/** 题意：CackingCodingInterview上的叠罗汉问题，leetcode上的沙皇问题* 将最长递增子序列扩展到了二维*//** solutions : 时间复杂度O(n+n*log(n)+n*log(n)) 也就是n*log(n)* 目标是将二维数组抽象成一维数组，再利用解最长递增子序列的解法来解决这个问题* 首先我们需要将二维数组排序，排序策略是先按照array[i][0]排序，在array[i][0]相同的情况下，按照array[i][1]倒序排序* 为了排序我们可将array抽象成一个类A，二元组的两个元素分别是这个类A的两个成员变量* 将array转化为A的数组，给A自实现一个排序函数（按照上面的排序策略）* 接下来就转化为了最长递增子序列问题，将类A数组的h成员变量看作之前的一维数组就好了*/public class Leetcode354 {public int maxEnvelopes(int envelopes) {if (envelopes.length < 2)return envelopes.length;A a = new A[envelopes.length];for (int i = 0; i < envelopes.length; i++) {a[i] = new A(envelopes[i][0],envelopes[i][1]);}Arrays.sort(a,new Comparator{@Overridepublic int compare(A a, A b) {if (a.w > b.w) {return 1;} else if (a.w < b.w) {return -1;} else {if (b.h > a.h) {return 1;} else if (b.h < a.h) {return -1;} elsereturn 0;}}});int maxLen = 0;int h = new int[envelopes.length];h[0] = a[0].h;for (int i = 1; i < envelopes.length; i++) {int pos = getPos(h, 0, maxLen - 1,a[i].h);if (pos == -1) {h[maxLen++] = a[i].h;} elseh[pos] = a[i].h;}return maxLen;}public int getPos(int h, int left, int right, int num) {if (left <= right) {int mid = left + (right - left) / 2;if (h[mid] >= num) {int pos = getPos(h, left, mid – 1, num);if (pos == -1)return mid;return pos;} else {return getPos(h, mid + 1, right, num);}}return -1;}public static void main(String args){Leetcode354 l = new Leetcode354;int envelopes = {{5,1},{6,4},{6,7},{2,3}};l.maxEnvelopes(envelopes);}}class A {int w = 0;int h = 0;public A(int w, int h) {this.w = w;this.h = h;}}View Code 第二题。