Day 7 | 454. 4Sum II | 383. Ransom Note | 15. 3Sum | 18. 4Sum

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本文精选了LeetCode上的多个经典算法题目，包括两数之和、三数之和、四数之和等，并提供了详细的解决方案及思路分析。

454. 4Sum II

Question Link

Solution:

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer, Integer> map = new HashMap<>();
        int count = 0; // count the present time of a+b+c+d = 0;
        for(int i : nums1){
            for(int j : nums2){
                int tmp = i + j;
                if(map.containsKey(tmp))
                	//use the key to store the sum of a and b
                	//use the value to keep the present time of the sum of a and b.
                    map.put(tmp, map.get(tmp)+1);
                else
                    map.put(tmp, 1);
            }
        }
        for(int i : nums3){
            for(int j : nums4){
                int tmp = i + j;
                if(map.containsKey(0-tmp))
                    count += map.get(0-tmp);
            }
        }
        return count;
    }
}

Thought:

Define a map, use the key to store the sum of a and b, and use the value to keep the present time of the sum of a and b
Define a count for counting the present time of a+b+c+d = 0
Iterate the nums3 and nums4, calculate the sum of c and d, that is tmp
If the map contains the key 0-tmp, use the count to pluse the value corresponding to the key in the map

383. Ransom Note

Question Link

Solution:

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] record = new int[26];
        for(char m : magazine.toCharArray()){
            record[m - 'a']++;
        }
        for(char r : ransomNote.toCharArray()){
            record[r - 'a']--;
        }
        
        for(int i : record){
            if(i < 0)
                return false;
        }
        return true;
    }
}

Thought:

Define an int array record for counting the presenting times of 26 characters
If the negative number exists, it demonstrates that at least a character exists in ransomNote but not in magazine.

15. 3Sum

Question Link

Solution:

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();       
        // 排序
        Arrays.sort(nums);
        for(int i = 0; i < nums.length; i++){
            // 如果第一个数a大于0，就不可能凑成三元组了
            if(nums[i] > 0)
                break;
            // 对a去重
            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)
                    right--;
                else if(sum < 0)
                    left++;
                else{
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    // 对b、c去重
                    while(left < right && nums[left] == nums[left+1]) left++;
                    while(left < right && nums[right] == nums[right-1]) right--;
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Thought:

Sort the array because we adopt the Double Pointer Method
After sorting, if the first element is greater than 0, other parts must be greater than 0. So we can do break directly.
Deduplicate a
According to the sum of three elements, adjusts the b and c until the sum equal to 0
Don’t forget to deduplicate b and c

18. 4Sum

Question Link

Solution:

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++){
            if(nums[i] > 0 && nums[i]>target)
                break;
            // deduplicate a
            if(i > 0 && nums[i] == nums[i-1])
                continue;
            for(int j = i+1; j < nums.length; j++){
                // deduplicate b
                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)
                        right--;
                    else if(sum < target)
                        left++;
                    else {
                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        // deduplicate c and d
                        while(left < right && nums[left]==nums[left+1]) left++;
                        while(left < right && nums[right]==nums[right-1])right--;

                        left++;
                        right--;
                    }
                }
            }
        }
        return result;
    }
}

Thought:

If the first element is greater than 0 and target, other parts must be greater than 0, and the sum must be greater than target. So we can do break directly.