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java_15.java
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java_15.java
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package leetcode;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
/**
* leetcode_15
*
* 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] ]
*
* @author rwxn
*
*/
public class java_15 {
List<List<Integer>> list = new ArrayList<List<Integer>>();
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {// 一层循环
if (nums[i] > 0) {
break;
}
if (nums[i] == 0 && i + 2 < nums.length) {
if (nums[i + 2] == 0) {
List<Integer> triplet = new ArrayList<Integer>();
Collections.addAll(triplet, nums[i], nums[i + 1], nums[i + 2]);
if (!list.contains(triplet)) {
list.add(triplet);
}
break;// 有三个元素为0,添加组合后跳出
}
break;
}
if (i >= 1 && nums[i] == nums[i - 1]) {
continue;
}
find_Twosum_equal_k(nums, i + 1, nums.length - 1, -nums[i]);
}
return list;
}
public void find_Twosum_equal_k(int[] num, int start, int real, int k) {
// 注意要找到所有可能情况,无重复数对
int j = start - 1;
while (start < real) {
if (num[start] + num[real] > k) {
real--;
} else if (num[start] + num[real] < k ) {
start++;
} else {
while (num[real] == num[real - 1] && start < real-1) {
real--;
}
while (num[start] == num[start + 1] && start+1 < real) {
start++;
}
List<Integer> triplet = new ArrayList<Integer>();
Collections.addAll(triplet, num[j], num[start], num[real]);
if (!list.contains(triplet)) {
list.add(triplet);
}
start++;
real--;
}
}
}
}