/* package whatever; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Ideone { public static void main (String[] args) throws ja…

For example we have: ["p", "r", "e", "f", "e", "t", " ", "m", "a", "k", "e", " ", "p", "r", "a", "t&…

Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 代码如下: public class Solution { public int reverse(int n) { long sum=0; int flag=1; if(n<0) { flag=-1; n=n*(-1); } try{ String s=Integer.toString(n); s=new Strin…

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 在一个反转了的排好序的数组中,找出最小的数 可以直接寻找. public class Solutio…

public class Solution { public int reverse(int x) { StringBuffer sb = new StringBuffer(x+"").reverse(); if(sb.charAt(sb.length()-1)=='-') { sb=new StringBuffer(sb.substring(0,sb.length()-1)); } int res=0; try { res= Integer.parseInt(sb.toString(…

反转链表,该链表为单链表. head 节点指向的是头节点. 最简单的方法,就是建一个新链表,将原来链表的节点一个个找到,并且使用头插法插入新链表.时间复杂度也就是O(n),空间复杂度就需要定义2个节点. 一个节点prev指向新的链表头,另一个节点temp用来获取原始链表的数据. public class Solution { /** * @param head: n * @return: The new head of reversed linked list. */ public ListNo…

public int reverse(int x) { long res = 0; while (x != 0){ res = res* 10 + x % 10; x /= 10; } if(res >= Integer.MAX_VALUE || res < Integer.MIN_VALUE) return 0; return (int)res; }…

数组颠倒算法 #include <iostream> #include <iterator> using namespace std; void reverse(int* A, int lo, int hi) { if (lo < hi) { swap(A[lo], A[hi]); reverse(A, ++lo, --hi); } } void reverse(int* A,int n) { reverse(A, , n-); } int main() { ] = { ,…

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题目: 假设按照升序排序的数组在预先未知的某个点上进行了旋转. ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] ). 搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 . 你可以假设数组中不存在重复的元素. 你的算法时间复杂度必须是 O(log n) 级别. 示例 1: 输入: nums = [4,5,6,7,0,1,2], target = 0 输出: 4 示例 2: 输入: nums = [4,5,6,7,0,1,2],…