Leetcode 3Sum 高效解法[Java]

以下是经过探索后得到的Leetcode上3Sum的高效解法, 当时能跑100.00%

class Solution {
    List<List<Integer>> res = new ArrayList<>();
	public List<List<Integer>> threeSum(int[] nums) {
		int len = nums.length;
		if (len < 3)
			return res;
        
        Arrays.sort(nums);  //sort the array first
        
        int zeroCount; //the appearing times of 0
        int lastNeg = Arrays.binarySearch(nums, 0); //search the position of 0; it also means the position of the last negative number in array
        int firstPos = lastNeg; //the position of the first positive number in array
        if(lastNeg < 0){    //0 not found
            zeroCount = 0;
            lastNeg = -(lastNeg + 1) - 1;//see the Java api
            firstPos = lastNeg + 1;
        }
        else{               //found
            while(lastNeg > -1 && nums[lastNeg] == 0) //skip all 0
                lastNeg--;
            while(firstPos < len && nums[firstPos] == 0)
                firstPos++;
            zeroCount = firstPos - lastNeg - 1;
        }


        int min;
        int max;
        int[] hash;
		min = nums[0];
		max = nums[len - 1];
		max = Math.max(Math.abs(max), Math.abs(min)); //to allocate enough space to avoid check in if statement
		min = -max;                                
		hash = new int[max - min + 1];
		for (int v : nums) { //hash and count appearing times of every num
			hash[v - min]++;
		}
        
		if (zeroCount >= 3) { // (0 appears 3 times at least)
			addTriplets(0, 0, 0);
		}
		if (zeroCount > 0 ) { // (0 appears 1 times at least)
			for (int i = firstPos; i < len; i++) { //traverse all the positive numbers to see whether there is a negative number whose absolute value equals to the positive number 
                if(i > firstPos && nums[i] == nums[i - 1]) //skip the same elements
                    continue;
                if (hash[-nums[i] - min] > 0) 
					addTriplets(0, nums[i], -nums[i]);
			}
		}


		// one positive number and two negetive numbers 
		for (int i = firstPos; i < len; i++) { //traverse all the positive numbers to find whether there are two negative numbers to make the 3 numbers added up to 0
            if(i > firstPos && nums[i] == nums[i - 1]) //skip the same elements
                    continue;
            int half;   //we can traverse only half of the positive numbers
            if(nums[i] % 2 != 0)
                half = -(nums[i] / 2 + 1);
            else{
                half = -(nums[i] / 2);
                if(hash[half - min] > 1)
                    addTriplets(nums[i], half, half);
            }
            for(int j = lastNeg; j > -1 && nums[j] > half; j--){
                if(j < lastNeg && nums[j] == nums[j + 1])
                    continue;
                if(hash[(-nums[i] - nums[j]) - min] > 0)
                    addTriplets(nums[i], nums[j], -nums[i] - nums[j]);
            }
        }
        
        // one negative number and two positive numbers 
		for (int i = lastNeg; i > -1; i--) { //traverse all the negative numbers to find whether there are two positive numbers to make the 3 numbers added up to 0
            if(i < lastNeg && nums[i] == nums[i + 1])//skip the same elements
                    continue;
            int half; //we can traverse only half of the negative numbers
            if(nums[i] % 2 != 0)
                half = -(nums[i] / 2 - 1);
            else{
                half = -(nums[i] / 2);
                if(hash[half - min] > 1)
                    addTriplets(nums[i], half, half);
            }
            for(int j = firstPos; j < len && nums[j] < half; j++){
                if(j > firstPos && nums[j] == nums[j - 1])
                    continue;
                if(hash[(-nums[i] - nums[j]) - min] > 0)
                    addTriplets(nums[i], nums[j], -nums[i] - nums[j]);
            }
        }
		return res;
	}


	public void addTriplets(int a, int b, int c) {
		List<Integer> triplets = new ArrayList<>(3);
		triplets.add(a);
		triplets.add(b);
		triplets.add(c);
		res.add(triplets);
	}
}
//MaplePC

水了一波顺丰科技的网上笔试之后, 最近一直在刷Leetcode, 也刷了有三十多道Array章节下面的题目了吧. 做题的时候习惯从最容易想到的方法开始探索更优的解法, 同时也有参考别人的代码汲取灵感, 并把这个过程形成的代码都记录在了GitHub上面, 以下是传送门:  https://github.com/MaplePc/LeetCode-Notes