代码随想录算法训练营20期|第七天|哈希表part02|454.四数相加II ● 383. 赎金信 ● 15. 三数之和 ● 18. 四数之和 ● 总结

454.四数相加II 

比较巧思的解法，先把nums1 和nums2的数两两相加，并存储sum和次数

再在nums3和nums4里找对应和sum和为0的数值i,j

Time: N^2

Space:N^2, 最坏情况下A和B的值各不相同，相加产生的数字个数为 n^2

class Solution {public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {Map<Integer, Integer> map = new HashMap<>();int res = 0;for (int i : nums1) {for (int j : nums2) {int sum = i + j;map.put(sum, map.getOrDefault(sum, 0) + 1);}}for (int i : nums3) {for (int j : nums4) {res += map.getOrDefault(0 - i - j, 0);}}return res;}
}

•  383. 赎金信 

先遍历长的

class Solution {public boolean canConstruct(String ransomNote, String magazine) {if (ransomNote.length() > magazine.length()) return false;int[] count = new int[26];for (char c : magazine.toCharArray()) {count[c - 'a']++;}for (char c : ransomNote.toCharArray()) {count[c - 'a']--;}for (int n : count) {if (n < 0) return false;}return true;}
}

•  15. 三数之和 

class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> res = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {if (nums[i] > 0) return res;if (i > 0 && nums[i] == nums[i - 1]) continue;int left = i + 1;int right = nums.length - 1;while (left < right) {int sum = nums[i] + nums[left] + nums[right];if (sum < 0) {left++;} else if (sum > 0) {right--;} else {res.add(Arrays.asList(nums[i], nums[left], nums[right]));while (left < right && nums[left] == nums[left + 1]) left++;while (left < right && nums[right] == nums[right-1]) right--;left++;right--;}}}return res;}
}

•  18. 四数之和 

在三数之和外面再套一层

class Solution {public List<List<Integer>> fourSum(int[] nums, int target) {List<List<Integer>> res = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {if (nums[i] > 0 && nums[i] > target) return res;if (i > 0 && nums[i] == nums[i - 1]) continue;for (int j = i + 1; j < nums.length; j++) {if (j > i + 1 && nums[j] == nums[j - 1]) continue;int left = j + 1;int right = nums.length - 1;while (left < right) {int sum = nums[i] + nums[j] + nums[left] + nums[right];if (sum < target) {left++;} else if (sum > target) {right--;} else {res.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));while (left < right && nums[left] == nums[left + 1]) left++;while (left < right && nums[right] == nums[right - 1]) right--;left++;right--;}}}}return res;}
}

•  总结