和为0的三数
和为0的三数
题目
给定一个未排序数数组,找出其中所有相加为0的不同的三元组。
示例1:
输入:[-3, 0, 1, 2, -1, 1, -2]
输出:[-3, 1, 2], [-2, 0, 2], [-2, 1, 1], [-1, 0, 1]
解释:上面4个集合的数的和为0
示例2:
输入:[-5, 2, -1, -2, 3]
输出:[-5, 2, 3], [-2, -1, 3]
解释:上面2个集合的数的和为0
解答1:
import java.util.*;
class TripletSumToZero {
public static List<List<Integer>> searchTriplets(int[] arr) {
Arrays.sort(arr);
List<List<Integer>> triplets = new ArrayList<>();
for (int i = 0; i < arr.length - 2; i++) {
if (i > 0 && arr[i] == arr[i - 1]) //跳过相同的元素去避免重复的三元组
continue;
searchPair(arr, -arr[i], i + 1, triplets);
}
return triplets;
}
private static void searchPair(int[] arr, int targetSum, int left, List<List<Integer>> triplets) {
int right = arr.length - 1;
while (left < right) {
int currentSum = arr[left] + arr[right];
if (currentSum == targetSum) { //找到三元组
triplets.add(Arrays.asList(-targetSum, arr[left], arr[right]));
left++;
right--;
//此处需要left<right是因为下面需要left++,而left++之前一定要检查left是否小于right
while (left < right && arr[left] == arr[left - 1])
left++; //跳过相同的元素避免相同的三元组
while (left < right && arr[right] == arr[right + 1])
right--; //跳过相同的元素避免相同的三元组
} else if (targetSum > currentSum)
left++; //我们将left指针右移来得到一个更大的数,从而得到更大的currentSum
else
right--; //我们将right指针左移来得到一个更小的数,从而得到更小的currentSum
}
}
public static void main(String[] args) {
System.out.println(TripletSumToZero.searchTriplets(new int[] { -3, 0, 1, 2, -1, 1, -2 }));
System.out.println(TripletSumToZero.searchTriplets(new int[] { -5, 2, -1, -2, 3 }));
}
}
function searchTriplets(arr) {
arr.sort((a, b) => a - b);
const triplets = [];
for (let i = 0; i < arr.length - 2; i++) {
if (i > 0 && arr[i] === arr[i - 1])
continue;
searchPair(arr, -arr[i], i + 1, triplets);
}
return triplets;
}
function searchPair(arr, targetSum, left, triplets) {
let right = arr.length - 1;
while (left < right) {
const currentSum = arr[left] + arr[right];
if (currentSum === targetSum) {
triplets.push([-targetSum, arr[left], arr[right]]);
left++;
right--;
while (left < right && arr[left] === arr[left - 1])
left++;
while (left < right && arr[right] === arr[right + 1])
right--;
} else if (targetSum > currentSum) {
left++;
} else {
right--;
}
}
}
console.log(searchTriplets([-3, 0, 1, 2, -1, 1, -2]));
console.log(searchTriplets([-5, 2, -1, -2, 3]));
def searchTriplets(arr):
arr.sort()
triplets = []
for i in range(len(arr) - 2):
if i > 0 and arr[i] == arr[i - 1]:
continue
searchPair(arr, -arr[i], i + 1, triplets)
return triplets
def searchPair(arr, targetSum, left, triplets):
right = len(arr) - 1
while left < right:
currentSum = arr[left] + arr[right]
if currentSum == targetSum:
triplets.append([-targetSum, arr[left], arr[right]])
left += 1
right -= 1
while left < right and arr[left] == arr[left - 1]:
left += 1
while left < right and arr[right] == arr[right + 1]:
right -= 1
elif targetSum > currentSum:
left += 1
else:
right -= 1
print(searchTriplets([-3, 0, 1, 2, -1, 1, -2]))
print(searchTriplets([-5, 2, -1, -2, 3]))
package main
import (
"fmt"
"sort"
)
func searchTriplets(arr []int) [][]int {
sort.Ints(arr)
triplets := [][]int{}
for i := 0; i < len(arr)-2; i++ {
if i > 0 && arr[i] == arr[i-1] {
continue
}
searchPair(arr, -arr[i], i+1, &triplets)
}
return triplets
}
func searchPair(arr []int, targetSum int, left int, triplets *[][]int) {
right := len(arr) - 1
for left < right {
currentSum := arr[left] + arr[right]
if currentSum == targetSum {
*triplets = append(*triplets, []int{-targetSum, arr[left], arr[right]})
left++
right--
for left < right && arr[left] == arr[left-1] {
left++
}
for left < right && arr[right] == arr[right+1] {
right--
}
} else if targetSum > currentSum {
left++
} else {
right--
}
}
}
func main() {
fmt.Println(searchTriplets([]int{-3, 0, 1, 2, -1, 1, -2}))
fmt.Println(searchTriplets([]int{-5, 2, -1, -2, 3}))
}