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3Sum.java
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64 lines (54 loc) · 2.14 KB
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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)
*/
public class Solution {
public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
Arrays.sort(num);
for (int i = 0; i < num.length; i++) {
if(i > 0 && num[i] == num[i - 1]) {
continue;
}
for (int j = i + 1; j < num.length; j++) {
int cur = num[i] + num[j];
int remainder = 0 - cur;
if (findTarget(num, j + 1, num.length - 1, remainder)) {
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(num[i]);
temp.add(num[j]);
temp.add(remainder);
if (!result.contains(temp)) {
result.add(temp);
}
}
}
}
return result;
}
public boolean findTarget(int[] num, int startIndex, int endIndex, int target) {
if (startIndex < 0 || endIndex >= num.length || startIndex > endIndex || target < num[startIndex] || target > num[endIndex]) {
return false;
}
while (startIndex <= endIndex) {
int midIndex = (startIndex + endIndex)/2;
int cur = num[midIndex];
if (cur == target) {
return true;
} else if (cur < target) {
startIndex = midIndex + 1;
} else {
endIndex = midIndex - 1;
}
}
return false;
}
}