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