1. Container With Most Water
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
思路：可以遍历线条所有两两组合的情况来解，但是复杂度回达到O(n2)，超时。采用另一种方法，从数组两端的点a1和an开始考虑，先计算a1和an所构成的Area，也就是a1和an中最短的那根线乘以a1到an x轴上的距离。然后开始往中间考虑，途中记录最大的Area。具体做法是，以a1 和 an为例，同时往左往右考虑，抛弃其中较短的线，因为对于a1和an中的较短者来说，x轴上的距离现在是最大了，也就是说不管剩下的那些线条比这个较短者（a1或者an）长还是短（因为装多少水取决于短板），所构成的Area都不会比原来a1和an构成的Area要大，所以在这个求最大Area的过程中可以直接将这个较短者剔除，移动至a2或者a(n-1)。再重复此步骤，每一步得到的Area都要和最大Area比较以实时更新。
详细见：https://leetcode.com/problems/container-with-most-water/solution/
import java.util.*;
public class LeetCode{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();
int[] nums=new int[input.length()];
for(int i=0;i<input.length();i++){
nums[i]=input.charAt(i)-'0';
}
}
public static int maxArea(int[] height) {
int maxarea = 0, l = 0, r = height.length - 1;
while (l < r) {
maxarea = Math.max(maxarea, Math.min(height[l], height[r]) * (r - l));
if (height[l] < height[r])
l++;
else
r--;
}
return maxarea;
}
}
2. 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
解析：模仿3Sum那题的思路即可。
import java.util.*;
public class LeetCode{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();
int n=sc.nextInt();
String[] str=input.split(",| ");
int[] nums=new int[str.length];
for(int i=0;i<str.length;i++){
// Integer.ParseInt() 方法将String转为int
nums[i]=Integer.parseInt(str[i]);
}
System.out.println(closet(nums,n));
}
public static int closet(int[] nums,int target) {
Arrays.sort(nums);
// Integer.MAX_VALUE
int closet=Integer.MAX_VALUE;
int closet_sum=0;
for(int i=0;i<nums.length-2;i++){
int l=i+1,r=nums.length-1;
while(l<r){
int sum=nums[i]+nums[l]+nums[r];
// 利用 Math.abs()方法求绝对值
if(Math.abs(sum-target)<closet){
closet=Math.abs(sum-target);
closet_sum=sum;
}
if(sum-target>0)
r--;
else
l++;
}
}
return closet_sum;
}
}
3. Letter Combinations of a Phone Number
Given a digit string, return all possible letter combinations that the number could represent. A mapping of digit to letters (just like on the telephone buttons) is given below.
Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
思路：利用先进先出队列构造所有可能的组合，一开始进队列的应该是一个字母，进完后出队，每个与另一组中的字母组合成2个，进队，进完后再出队，与另一组中的字母组合构成3个，进队再出队，以此类推。
import java.util.*;
public class LeetCode{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
String digits=sc.nextLine();
List<String> letters=letterCombinations(digits);
for(String s:letters)
System.out.println(s);
}
public static List<String> letterCombinations(String digits) {
// LinkedList 实现了 Queue 接口和 List 接口，能当作队列来用
LinkedList<String> letters=new LinkedList<String>();
// 数组下标对应于输入的数字，值是数字对应的键盘上的字母
String[] mapping=new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
// 这里先往队列添个东西，因为空队列 remove 方法会抛异常
letters.add("");
for(int i=0;i<digits.length();i++){
int x=digits.charAt(i)-'0';
// 这里是个较难理解的点，i(digits的char索引，从0开始)和队列中目前的字母组合的长度(也就是1个字母还是2个字母或者0个字母空队列的情况)对应
while(letters.peek().length()==i){
String s=letters.remove();
for(char c:mapping[x].toCharArray()){
letters.add(s+c);
}
}
}
return letters;
}
}