LeetCode 15. 三数之和
2023-12-16 14:51:56
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
解法思路:
// a + b = -c (target)
// 1. 暴力求解:三重循环 O(n^3)
// 2. hash表来记录 a, b, a+b = -c
// 3. 双指针:左右下标往中间推进
?法一:
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// 暴力求解:三重循环,O(n^3)
Set set = new HashSet<>();
for (int i = 0; i < nums.length - 2; i++) {
for (int j = i + 1; j < nums.length - 1; j++) {
for (int k = j + 1; k < nums.length; k++) {
if (nums[i] + nums[j] + nums[k] == 0) {
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[k]);
Collections.sort(list);
set.add(list);
}
}
}
}
return new ArrayList<>(set);
}
}
// 小小优化
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// 暴力求解:三重循环,O(n^3)
Arrays.sort(nums);
Set set = new HashSet<>();
for (int i = 0; i < nums.length - 2; i++) {
for (int j = i + 1; j < nums.length - 1; j++) {
for (int k = j + 1; k < nums.length; k++) {
if (nums[i] + nums[j] + nums[k] == 0) {
set.add(Arrays.asList(nums[i], nums[j], nums[k])); }
}
}
}
return new ArrayList<>(set);
}
}
法二:
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// Time: O(n^2)
Arrays.sort(nums);
List<List<Integer>> res = new ArrayList<>();
if (nums.length < 3 || nums[0] > 0) {
return res;
}
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int i = 0; i < nums.length - 2; i++) {
if (nums[i] > 0) {
break;
}
for (int j = i + 1; j < nums.length - 1; j++) {
int target = -1 * (nums[i] + nums[j]);
if (map.containsKey(target) && map.get(target) > j) {
res.add(Arrays.asList(nums[i], nums[j], target));
}
j = map.get(nums[j]);
}
i = map.get(nums[i]);
}
return res;
}
}
法三:
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// Time: O(n^2)
Arrays.sort(nums);
List<List<Integer>> list = new ArrayList<>();
if (nums.length < 3 || nums[0] > 0) {
return list;
}
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0) {
break;
}
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1, right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum > 0) {
right--;
} else if (sum < 0) {
left++;
} else {
list.add(Arrays.asList(nums[i], nums[left], nums[right]));
int lastLeftCur = nums[left];
int lastRightCur = nums[right];
while (left < right && nums[left] == lastLeftCur) {
left++;
}
while (left < right && nums[right] == lastRightCur) {
right--;
}
}
}
}
return list;
}
}
// 另一个版本
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
// Time: O(n^2)
Arrays.sort(nums);
List<List<Integer>> res = new ArrayList<>();
for (int k = 0; k < nums.length - 2; ++k) {
if (nums[k] > 0) break;
if (k > 0 && nums[k] == nums[k - 1]) continue;
int i = k + 1;
int j = nums.length - 1;
while (i < j) {
int sum = nums[k] + nums[i] + nums[j];
if (sum < 0) {
while (i < j && nums[i] == nums[++i]);
} else if (sum > 0) {
while (i < j && nums[j] == nums[--j]);
} else {
res.add(new ArrayList<>(Arrays.asList(nums[k], nums[i], nums[j])));
while (i < j && nums[i] == nums[++i]);
while (i < j && nums[j] == nums[--j]);
}
}
}
return res;
}
}
文章来源:https://blog.csdn.net/qq_38304915/article/details/135031526
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