[LeetCode] 259. 3Sum Smaller 三数之和较小值
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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(n2) 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] 16. 3Sum Closest 最近三数之和
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[LeetCode] 259. 3Sum Smaller 三数之和较小值