454. 4Sum II
Given four integer arrays nums1
, nums2
, nums3
, and nums4
all of length n
, return the number of tuples (i, j, k, l)
such that:
0 <= i, j, k, l < n
nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0
Example 1:
Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
Example 2:
Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1
Constraints:
n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228
首先定义 一个map,key放a和b两数之和,value 放a和b两数之和出现的次数。
遍历大A和大B数组,统计两个数组元素之和,和出现的次数,放到map中。
定义int变量count,用来统计 a+b+c+d = 0 出现的次数。
在遍历大C和大D数组,找到如果 0-(c+d) 在map中出现过的话,就用count把map中key对应的value也就是出现次数统计出来。
最后返回统计值 count 就可以了
class Solution {
public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
Map<Integer, Integer> map = new HashMap<>();
int temp;
int res = 0;
//统计两个数组中的元素之和,同时统计出现的次数,放入map
for (int i : nums1) {
for (int j : nums2) {
temp = i + j;
if (map.containsKey(temp)) {
map.put(temp, map.get(temp) + 1);
} else {
map.put(temp, 1);
}
}
}
//统计剩余的两个元素的和,在map中找是否存在相加为0的情况,同时记录次数
for (int i : nums3) {
for (int j : nums4) {
temp = i + j;
if (map.containsKey(0 - temp)) {
res += map.get(0 - temp);
}
}
}
return res;
}
}
Time Complexity:O(n^2)
Space Complexity:O(n)
For Future References
题目链接:https://leetcode.com/problems/4sum-ii/
文章讲解: https://programmercarl.com/0454.四数相加II.html
视频讲解:https://www.bilibili.com/video/BV1Md4y1Q7Yh/
383. Ransom Note
Given two strings ransomNote
and magazine
, return true
if ransomNote
can be constructed by using the letters from magazine
and false
otherwise.
Each letter in magazine
can only be used once in ransomNote
.
Example 1:
Input: ransomNote = "a", magazine = "b"
Output: false
Example 2:
Input: ransomNote = "aa", magazine = "ab"
Output: false
Example 3:
Input: ransomNote = "aa", magazine = "aab"
Output: true
Constraints:
1 <= ransomNote.length, magazine.length <= 105
ransomNote
andmagazine
consist of lowercase English letters.
字符串a能否组成字符串b,而不用管字符串b 能不能组成字符串a。
-
第一点“为了不暴露赎金信字迹,要从杂志上搜索各个需要的字母,组成单词来表达意思” 这里说明杂志里面的字母不可重复使用。
-
第二点 “你可以假设两个字符串均只含有小写字母。” 说明只有小写字母,这一点很重要
暴力解法
两层for循环,不断去寻找
Hash table
因为题目所只有小写字母,那可以采用空间换取时间的哈希策略, 用一个长度为26的数组还记录magazine里字母出现的次数。然后再用ransomNote去验证这个数组是否包含了ransomNote所需要的所有字母。
如果数组中存在负数,说明ransomNote字符串总存在magazine中没有的字符
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
// 定义一个哈希映射数组
int[] record = new int[26];
// 遍历
for(char c : magazine.toCharArray()){
record[c - 'a'] += 1;
}
for(char c : ransomNote.toCharArray()){
record[c - 'a'] -= 1;
}
// 如果数组中存在负数,说明ransomNote字符串总存在magazine中没有的字符
for(int i : record){
if(i < 0){
return false;
}
}
return true;
}
}
Time Complexity:O(n)
Space Complexity:O(1)
For Future References
题目链接:https://leetcode.com/problems/ransom-note/
文章讲解: https://programmercarl.com/0383.赎金信.html
15. 3Sum
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]]
such that i != j
, i != k
, and j != k
, and nums[i] + nums[j] + nums[k] == 0
.
Notice that the solution set must not contain duplicate triplets.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
Constraints:
3 <= nums.length <= 3000
105 <= nums[i] <= 105
Two Pointers
首先将数组排序,然后有一层for循环,i从下标0的地方开始,同时定一个下标left 定义在i+1的位置上,定义下标right 在数组结尾的位置上。
依然还是在数组中找到 abc 使得a + b +c =0,我们这里相当于 a = nums[i],b = nums[left],c = nums[right]。
-
如果nums[i] + nums[left] + nums[right] > 0 就说明 此时三数之和大了,因为数组是排序后了,所以right下标就应该向左移动,这样才能让三数之和小一些。
-
如果 nums[i] + nums[left] + nums[right] < 0 说明 此时 三数之和小了,left 就向右移动,才能让三数之和大一些,直到left与right相遇为止。
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0) {
return result;
}
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.length - 1;
while (right > left) {
int sum = nums[i] + nums[left] + nums[right];
if (sum > 0) {
right--;
} else if (sum < 0) {
left++;
} else {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
right--;
left++;
}
}
}
return result;
}
}
Time Complexity:O(n^2)
Space Complexity:O(n)
For Future References
题目链接:https://leetcode.com/problems/3sum/
文章讲解: https://programmercarl.com/0015.三数之和.html
视频讲解:https://www.bilibili.com/video/BV1GW4y127qo/
18. 4Sum
Given an array nums
of n
integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]]
such that:
0 <= a, b, c, d < n
a
,b
,c
, andd
are distinct.nums[a] + nums[b] + nums[c] + nums[d] == target
You may return the answer in any order.
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:
Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
Constraints:
1 <= nums.length <= 200
109 <= nums[i] <= 109
109 <= target <= 109
四数之和的双指针解法是两层for循环nums[k] + nums[i]为确定值,依然是循环内有left和right下标作为双指针,找出nums[k] + nums[i] + nums[left] + nums[right] == target的情况,三数之和的时间复杂度是O(n2),四数之和的时间复杂度是O(n3) 。
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
// nums[i] > target 直接返回, 剪枝操作
if (nums[i] > 0 && nums[i] > target) {
return result;
}
if (i > 0 && nums[i - 1] == nums[i]) {
continue;
}
for (int j = i + 1; j < nums.length; j++) {
if (j > i + 1 && nums[j - 1] == nums[j]) {
continue;
}
int left = j + 1;
int right = nums.length - 1;
while (right > left) {
long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
if (sum > target) {
right--;
} else if (sum < target) {
left++;
} else {
result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
left++;
right--;
}
}
}
}
return result;
}
}
Time Complexity:O(n^3)
Space Complexity:O(n)
For Future References
题目链接:https://leetcode.com/problems/4sum/
文章讲解: https://programmercarl.com/0018.四数之和.html
视频讲解:https://www.bilibili.com/video/BV1DS4y147US/
Summary
一般来说哈希表都是用来快速判断一个元素是否出现集合里。
接下来是常见的三种哈希结构:
- 数组
- set(集合)
- map(映射)