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  • leetcode — 3sum

    import java.util.*;
    
    /**
     * Source : https://oj.leetcode.com/problems/3sum/
     *
     * Created by lverpeng on 2017/7/10.
     *
     * 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:
     *
     * Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
     * 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)
     */
    public class SumEqualsZero {
    
        /**
         * 最简单的方法,计算出所有三个数和为0的情况
         *
         * @param s
         * @return
         */
        public Set<Integer[]> findThreeNum (int[] s) {
            Arrays.sort(s);
            System.out.println(Arrays.toString(s));
            Set<Integer[]> result = new HashSet<Integer[]>();
            if (s.length < 4) {
                return null;
            }
            for (int i = 0; i < s.length - 2; i++) {
                for (int j = i + 1; j < s.length - 1; j++) {
                    for (int k = j + 1; k < s.length; k++) {
                        if(s[i] + s[j] + s[k] == 0) {
                            Integer[] arr = {s[i], s[j], s[k]};
                            result.add(arr);
                        }
                    }
                }
            }
    
            return result;
        }
    
        /**
         * 可以转化为和twosum一样的问题,相当于是多个twosum问题
         * a + b = -c
         * 就是两个数的和是一个定值,针对每一种c的情况求出a、b
         *
         * @param s
         * @return
         */
        public Set<Integer[]> findThreeNum1 (int[] s) {
            Arrays.sort(s);
            Set<Integer[]> set = new HashSet<Integer[]>();
            for (int i = 0; i < s.length - 2; i++) {
                int total = -s[i];
                int left = i + 1;
                int right = s.length -1;
                while (left < right) {
                    if (s[left] + s[right] == total) {
                        Integer[] arr = {s[i], s[left], s[right]};
                        set.add(arr);
                        left ++;
                        right --;
                    } else if (s[left] + s[right] > total) {
                        while (left < right && s[left] + s[right] > total) {
                            right --;
                        }
                    } else {
                        while (left < right && s[left] + s[right] < total) {
                            left ++;
                        }
                    }
                }
            }
            return set;
        }
    
        public static void main(String[] args) {
            SumEqualsZero sumEqualsZero = new SumEqualsZero();
            int[] arr = {-1, 0 ,1, 2, -1, -4};
            printList(sumEqualsZero.findThreeNum(arr));
            printList(sumEqualsZero.findThreeNum1(arr));
        }
    
        public static void printList (Set<Integer[]> list) {
            for (Integer[] i : list) {
                System.out.println(Arrays.toString(i));
            }
        }
    
    }
    
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  • 原文地址:https://www.cnblogs.com/sunshine-2015/p/7342002.html
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