3sum, 3sum closest
[抄题]:
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.
[思维问题]:
[一句话思路]:
化成和为0 - nums[i]的两根指针2sum
[输入量特别大怎么办]:
[画图]:
[一刷]:
- 记得第一步就写排序
- corner case反正都是空的,直接return res自己就行。
- 防止重复判断时必须要写(i > 0 && nums[i] != nums[i - 1]),因为i - 1最后一位是0。
- 下次不要定义value了,因为value随指针的改变而不断变化。
- List<List<Integer>> res = new ArrayList<List<Integer>>(); 只有最右边是空的。
- 循环条件是for (int i = 0; i < nums.length; i++),上界还是i < nums.length
[总结]:
不要定义value
记得先排序
[复杂度]:
[英文数据结构,为什么不用别的数据结构]:
用linkedlist: 3sum的元素个数不确定,需要动态添加
[其他解法]:
[题目变变变]:
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> res = new LinkedList<>(); if (nums == null || nums.length < 3) {
return res;
} Arrays.sort(nums);//!!! for (int i = 0; i < nums.length; i++) {
int left = i + 1;
int right = nums.length - 1;
int sum = 0 - nums[i]; if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
while(left < right) {
if (nums[left] + nums[right] == sum) {
res.add(Arrays.asList(nums[i],nums[left],nums[right]));
while(left < right && nums[left] == nums[left + 1]) {
left++;
}
while(left < right && nums[right] == nums[right - 1]) {
right--;
}
left++;
right--;
}
else if (nums[left] + nums[right] < sum) {
left++;
}
else {
right--;
}
}
}
}
return res;
}
}
3sum closest
[抄题]:
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
[思维问题]:
用diff做要判断两边,很麻烦。
[一句话思路]:
用abs绝对值函数,直接就两边比较大小。
[输入量特别大怎么办]:
[画图]:
[一刷]:
用了指针的变量,要放在发生指针变化的循环体之内。
[总结]:
注意用了指针的变量
[复杂度]:
[英文数据结构,为什么不用别的数据结构]:
[其他解法]:
[Follow Up]:
[题目变变变]:
2sum closest 4...
class Solution {
public int threeSumClosest(int[] nums, int target) {
if (nums.length < 3 || nums == null) {
return -1;
} Arrays.sort(nums);
int bestSum = nums[0] + nums[1] + nums[2];
for(int i = 0; i < nums.length; i++) {
int left = i + 1;
int right = nums.length - 1; while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (Math.abs(sum - target) < Math.abs(bestSum - target)) {
bestSum = sum;
} if (sum == target) {
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 bestSum;
}
}
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