正文
[LeetCode] 259. 3Sum Smaller 三数之和较小值
提示：扫一扫查出行【扫一扫了解最新限行尾号】
复制提示
Given an array of
n
integers
nums
and a
target
, find the number of index triplets
i, j, k
with
0 <= i < j < k < n
that satisfy the condition
nums[i] + nums[j] + nums[k] < target
.
For example, given
nums
=
[-2, 0, 1, 3]
, and
target
= 2.
Return 2. Because there are two triplets which sums are less than 2:
[-2, 0, 1]
[-2, 0, 3]
Follow up:
Could you solve it in
O
(
n
2) runtime?
给一个整数数组nums和一个target，找出所有3个指针对应数字和小于target的组合的数量。
解法1: 暴力brute force, 找出所有组合的和，然后找出小于target的组合，O(n^3)
解法2: 双指针，类似3Sum的方法，但由于只需要求count，而不用求出每个组合，可以做到O(n^2)。还是先对数组排序，然后用2个指针前后夹逼，当i, lo, hi这个组合满足条件时，在[lo, hi]这个闭合区间内的所有组合也应该满足条件，所以可以直接count += hi - lo，然后lo++，增大三个值的和来继续尝试，假如不满足条件，则hi--来缩小三个值的和。
Java:
public class Solution {
public int threeSumSmaller(int[] nums, int target) {
if(nums == null || nums.length == 0)
return 0;
Arrays.sort(nums);
int count = 0;
for(int i = 0; i < nums.length - 2; i++) {
int lo = i + 1, hi = nums.length - 1;
while(lo < hi) {
if(nums[i] + nums[lo] + nums[hi] < target) {
count += hi - lo;
lo++;
} else {
hi--;
}
}
}
return count;
}
}
Python:
# Time: O(n^2)
# Space: O(1)
class Solution:
# @param {integer[]} nums
# @param {integer} target
# @return {integer}
def threeSumSmaller(self, nums, target):
nums.sort()
n = len(nums) count, k = 0, 2
while k < n:
i, j = 0, k - 1
while i < j: # Two Pointers, linear time.
if nums[i] + nums[j] + nums[k] >= target:
j -= 1
else:
count += j - i
i += 1
k += 1 return count
C++:
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
if (nums.size() < 3) return 0;
int res = 0, n = nums.size();
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 2; ++i) {
int left = i + 1, right = n - 1;
while (left < right) {
if (nums[i] + nums[left] + nums[right] < target) {
res += right - left;
++left;
} else {
--right;
}
}
}
return res;
}
};
类似题目：
[LeetCode] 15. 3Sum 三数之和
[LeetCode] 16. 3Sum Closest 最近三数之和
