Discuss：www.cnblogs.com/grandyang/p…
Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:
0 <= a, b, c, d < n a, b, c, and d are distinct. nums[a] + nums[b] + nums[c] + nums[d] == target You may return the answer in any order.
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:
Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
Constraints:
-
1 <= nums.length <= 200
-
-109 <= nums[i] <= 109
-
-109 <= target <= 109
解法一：
此题与 Three Sum 类似，就是多了一层循环，手动进行去重复处理。主要可以进行的有三个地方，首先在两个 for 循环下可以各放一个，因为一旦当前的数字跟上面处理过的数字相同了，那么找下来肯定还是重复的。之后就是当 sum 等于 target 的时候了，在将四个数字加入结果 res 之后，left 和 right 都需要去重复处理，分别向左右方向遍历即可。
class Solution {
fun fourSum(nums: IntArray, target: Int): List<List<Int>> {
if (nums.size < 4) {
return mutableListOf()
}
val result = mutableListOf<List<Int>>()
nums.sort()
for (i in 0..nums.size - 4) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue
}
for (j in i + 1..nums.size - 3) {
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue
}
var left = j + 1
var right = nums.size - 1
while (left < right) {
val sum = nums[i] + nums[j] + nums[left] + nums[right]
if (sum == target) {
val item = mutableListOf(nums[i], nums[j], nums[left], nums[right])
result.add(item)
while (left < right && nums[left] == nums[left + 1]) {
++left
}
while (left < right && nums[right] == nums[right - 1]) {
--right
}
++left
--right
} else if (sum < target) {
++left
} else {
--right
}
}
}
}
return result
}
}
