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) ```分析：
对数组进行排序；
start从0到n-1，对numstart求另外两个数，这里用mid和end；
mid指向start 1，q指向结尾。sum = numstart nummid numend；
利用加逼定理求解，终止条件是mid == end；
顺带去重
去重：
如果start到了start 1，numstart == numstart - 1，则求出的解肯定重复了；
如果mid ，nummid = nummid - 1，则求出的解肯定重复了。
代码语言：txt复制 public List<List<Integer>> threeSum(int[] nums) {
if (nums == null || nums.length < 3) {
return new ArrayList<List<Integer>>();
}
Set<List<Integer>> set = new HashSet<List<Integer>>();
Arrays.sort(nums);
for (int start = 0; start < nums.length; start ) {
if (start != 0 && nums[start - 1] == nums[start]) {
continue;
}
int mid = start 1, end = nums.length - 1;
while (mid < end) {
int sum = nums[start] nums[mid] nums[end];
if (sum == 0) {
List<Integer> tmp = new ArrayList<Integer>();
tmp.add(nums[start]);
tmp.add(nums[mid]);
tmp.add(nums[end]);
set.add(tmp);
while ( mid < end && nums[mid - 1] == nums[mid])
;
while (--end > mid && nums[end 1] == nums[end])
;
}
else if (sum < 0) {
mid ;
}
else {
end--;
}
}
}
return new ArrayList<List<Integer>>(set);
}
