算法导论:22页2.3-7 描述一个运行时间为O(nlogn)的算法,找出n个元素的S数组中是否存在两个元素相加等于给定x值 AC解: a=[1,3,6,7,9,15,29] def find2sumx(nums,x): nums.sort() le,ri=0,len(nums)-1 while le>=0 and ri<=len(nums) and le<ri: if nums[le]+nums[ri]<x: le+=1 elif nums[le]+nums[ri]>x:
package com.shb.java; /** * 将数组中的0放到数组的后边,然后原来的非零数的顺序不改变 * @author BIN * */ public class Demo2{ public static void main(String[] args){ getArray(new int[]{1,0,0,5,0}); } public static int[] getArray(int[] array){ int cur = 0; for(int i=0;i<array.leng
B. Powers of Two You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer xexists so that ai + aj = 2x). Input The first line contains the single positive integer n
题目: 给定一个数组candidates和一个目标值target,求出数组中相加结果为target的数字组合: 举例: For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: [[7],[2, 2, 3]] 从举例中可以看出,同一个数字可以使用多次,且结果是唯一的: 解题思路: 我个人感觉该题目一点也不难,其实也是一个递归的过程,当达到递归条件时,就将结果加入结果集: 首先题目没说给的数组有啥特
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. 这里题目要求找出出现次数超过n/2的元素. 可以先排序,
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. For example, given array S = [-1,