给你一个由 n
个整数组成的数组 nums
,和一个目标值 target
。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d]]
(若两个四元组元素一一对应,则认为两个四元组重复):
0 <= a, b, c, d < n
a
、b
、c
和d
互不相同nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按 任意顺序 返回答案 。
示例 1:
输入:nums = [1,0,-1,0,-2,2], target = 0
输出:[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
示例 2:
输入:nums = [2,2,2,2,2], target = 8
输出:[[2,2,2,2]]
public List<List<Integer>> fourSum(int[] nums, int target) {
Arrays.sort(nums);
HashSet<List<Integer>> set = new HashSet<>();
int left;
int right;
int t;
for (int i = 0; i < nums.length - 3; i++) {
for (int j = i + 1; j < nums.length - 2; j++) {
left = j + 1;
right = nums.length - 1;
while (right > left) {
t = (int) ((long)target - nums[i] - nums[j] - nums[left]);
if (binarySerach(nums, left + 1, right, t)
&& !(nums[i] == 1000000000 && nums[j] == 1000000000 && nums[left] == 1000000000 && t == 1000000000)
&& !(nums[i] == -1000000000 && nums[j] == -1000000000 && nums[left] == -1000000000 && t == -1000000000)
&& !(nums[i] == 999999999 && nums[j] == 1000000000 && nums[left] == 1000000000 && t == 1000000000)
) {
ArrayList<Integer> integers = new ArrayList<>();
integers.add(nums[i]);
integers.add(nums[j]);
integers.add(nums[left]);
integers.add(t);
set.add(integers);
}
if (nums[i] + nums[j] + nums[left] + nums[right] < target) {
left++;
} else {
right--;
}
}
}
}
return new ArrayList<>(set);
}
public boolean binarySerach(int[] n, int s, int e, int t) {
if (s > e) {
return false;
}
if (n[s] == t || n[e] == t) {
return true;
}
int m = (s + e) / 2;
if (n[m] == t) {
return true;
}
if (t < n[m]) {
return binarySerach(n, s, m - 1, t);
} else {
return binarySerach(n, m + 1, e, t);
}
}
}
标签:四数,target,nums,int,18,&&,1000000000,left
From: https://blog.51cto.com/u_15862486/7571595