3Sum_Leetcode_#15

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1.题目
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

2.解法
思路：先给nums排序，最大值nums[nums.length - 1]与最小值nums[0]之间的距离为gap，取一个数组countNum用来统计nums[i] - nums[0]出现的次数。以-nums[0]作为基准base，(-nums[0] > 0)，取x < base, y > base, x < z < y.找到符合x+y+z=3*base的条件的x,y,z.则x-base,y-base,z-base就是一个解。

时间复杂度为O(N^2)

public class Solution {    public List<List<Integer>> threeSum(int[] nums){        List arr = new ArrayList();        int size = nums.length;        if(size == 0){            return arr;        }        Arrays.sort(nums);        int gap = nums[size - 1] - nums[0] + 1;        int[] countNum = new int[gap];        for(int i = 0; i < size; i++){            countNum[nums[i] - nums[0]]++;        }        int base = -nums[0];        if(base < 0 || base >= countNum.length){            return arr;        }        if(countNum[base] >= 3){            arr.add(new int[]{0, 0, 0});        }        //x-base <0, y-base>0,  x<z<y        for(int x = 0; x < base; x = findNextLtBase(countNum, x)){            for(int y = countNum.length - 1; y > base; y = findNextGtBase(countNum, y)){                int z = 3 * base - x - y;                 if(z == x && countNum[x] >= 2 || z == y && countNum[y] >= 2 || z > x && z < y && countNum[z] >= 1){                    arr.add(new int[]{x - base, y - base, z - base});                }            }        }        return arr;    }    public int findNextLtBase(int[] countNum, int m){        for(++m; m < countNum.length; ++m){            if(countNum[m] > 0){                return m;            }        }        return m;    }    public int findNextGtBase(int[] countNum, int m){        for(--m; m >= 0; --m){            if(countNum[m] > 0){                return m;            }        }               return m;    }}

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