leetcode 17-18

leetcode 17 : Letter Combinations of a Phone Number

问题描述

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"].

分析

这道题是让我们求电话号码对应的所有字母组合，我们可以通过一个队列实现迭代，从头取出长度较短的字符串再加上一个字符添加到队尾。

JAVA代码

public List<String> letterCombinations(String digits) {    LinkedList<String> res = new LinkedList<>();    if(digits.length() == 0) {        return res;    }    //映射关系字符串数组    String[] m = new String[]{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    res.add("");    for(int i=0;i<digits.length();i++) {        int x = Character.getNumericValue(digits.charAt(i));        //取得最头的元素长度但不删除元素        while(res.peek().length() == i) {            //移除头元素并赋值给p            String p = res.remove();            for (char c : m[x].toCharArray()) {                //将每一个添加的字符连接到p上并添加到队列尾部                res.add(p+c);            }        }    }    return res;}

总结

善于灵活应用已有的数据结构的特性，这题就是我们在后面要用前面的数据，所以就利用先进先出的队列解决问题。

leetcode 18 ： 4Sum

问题描述

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[[-1,  0, 0, 1],[-2, -1, 1, 2],[-2,  0, 0, 2]]

分析

这个问题只需要在3Sum的基础上再加一层循环即可，实践复杂度为O(n^3)。

JAVA代码

public List<List<Integer>> fourSum(int[] nums, int target) {    Arrays.sort(nums);    List<List<Integer>> res = new LinkedList<>();    for(int f=0;f<nums.length-3;f++) {        int subtarget = target - nums[f];        if(f == 0 || nums[f] != nums[f-1]) {            for (int i = f + 1; i < nums.length - 2; i++) {                if (i == f+1 || nums[i] != nums[i - 1]) {                    int low = i + 1, high = nums.length - 1, sum = subtarget - nums[i];                    while (low < high) {                        if (sum == nums[low] + nums[high]) {                            res.add(Arrays.asList(nums[f], nums[i], nums[low], nums[high]));                            while (low < high && nums[low] == nums[low + 1]) {                                low++;                            }                            while (low < high && nums[high] == nums[high - 1]) {                                high--;                            }                            low++;                            high--;                        } else if (sum > nums[low] + nums[high]) {                            low++;                        } else {                            high--;                        }                    }                }            }        }    }    return res;}