LeetCode-3SUM(数组中三数之和为0)
时间:2015-02-07 17:33:41
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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)
第一想法肯定是三层循环遍历的,时间复杂度O(n^3),但是做过了了 Two Sum ,感觉瞬间可解,随用hashtable的方法:
public List<List<Integer>> threeSum1(int[] num) { List<List<Integer>> lists = new ArrayList<>(); HashMap<Integer, Integer> hash = new HashMap<>(); for (int i = 0; i < num.length; i++) { hash.put(num[i], i); } for (int i = 0; i < num.length; i++) { for (int j = i+1; j < num.length; j++) { if (hash.get(0-num[i]-num[j]) != null && num[i] <= num[j] && num[j] <= 0-num[i]-num[j]) { List<Integer> list = new ArrayList<>(); list.add(num[i]); list.add(num[j]); list.add(0-num[i]-num[j]); lists.add(list); } } } return lists; }
时间复杂度应该在O(n^2),但是不幸,直接超时!看着题意,需要在O(n)解决么?经分析,肯定不行,至少也要O(NlgN)。好吧,联想起排序的复杂度,所以,看是还是先把数组给排序了比较好控制。
先上代码:
public List<List<Integer>> threeSum(int[] num) { List<List<Integer>> lists = new ArrayList<>(); Arrays.sort(num); for (int i = 0; i < num.length - 2; i++) { if (i == 0 || (i > 0 && num[i] != num[i-1])) { int low = i+1, high = num.length-1, sum = 0-num[i]; while (low < high) { if (num[low]+num[high] == sum) { lists.add(Arrays.asList(num[i], num[low], num[high])); while (low < high && num[low] == num[low+1]) { low ++; } while (low <high && num[high] == num[high-1]) { high --; } low++; high--; } else if (num[low] + num[high] < sum) { low++; } else { high--; } } } } return lists; }
先控制一个变量a,线性遍历。此时只需找出b,c满足b+c=-a即可。由于是排序好的,剩下的可以用二分查找的思想找出b.c。于是求解(注意要排除重复解!!!!)
原文:http://blog.csdn.net/my_jobs/article/details/43603901
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