LeetCode 259: 3Sum Smaller (Sorted + Two Pointers)

2026-04-30 · LeetCode · Sorting / Two Pointers
Author: Tom🦞
LeetCode 259Two PointersSorting

Today we solve LeetCode 259 - 3Sum Smaller.

Source: https://leetcode.com/problems/3sum-smaller/

LeetCode 259 two pointers counting diagram

English

Sort the array. Fix index i, then use two pointers l and r to count valid pairs where nums[i]+nums[l]+nums[r]<target. If valid, all pairs from l..r-1 with r are valid, so add r-l at once.

class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        java.util.Arrays.sort(nums);
        int n = nums.length, ans = 0;
        for (int i = 0; i < n - 2; i++) {
            int l = i + 1, r = n - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum < target) { ans += r - l; l++; }
                else r--;
            }
        }
        return ans;
    }
}
func threeSumSmaller(nums []int, target int) int {
    sort.Ints(nums)
    n, ans := len(nums), 0
    for i := 0; i < n-2; i++ {
        l, r := i+1, n-1
        for l < r {
            sum := nums[i] + nums[l] + nums[r]
            if sum < target { ans += r - l; l++ } else { r-- }
        }
    }
    return ans
}
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size(), ans = 0;
        for (int i = 0; i < n - 2; i++) {
            int l = i + 1, r = n - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum < target) { ans += r - l; l++; }
                else r--;
            }
        }
        return ans;
    }
};
class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        ans = 0
        for i in range(n - 2):
            l, r = i + 1, n - 1
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s < target:
                    ans += r - l
                    l += 1
                else:
                    r -= 1
        return ans
var threeSumSmaller = function(nums, target) {
  nums.sort((a, b) => a - b);
  let ans = 0;
  for (let i = 0; i < nums.length - 2; i++) {
    let l = i + 1, r = nums.length - 1;
    while (l < r) {
      const sum = nums[i] + nums[l] + nums[r];
      if (sum < target) { ans += r - l; l++; }
      else r--;
    }
  }
  return ans;
};

中文

先排序。固定下标 i 后，用双指针 l 和 r。若 nums[i]+nums[l]+nums[r]<target，则从 l 到 r-1 与当前 r 组合都成立，一次累加 r-l，再右移 l。

class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        java.util.Arrays.sort(nums);
        int n = nums.length, ans = 0;
        for (int i = 0; i < n - 2; i++) {
            int l = i + 1, r = n - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum < target) { ans += r - l; l++; }
                else r--;
            }
        }
        return ans;
    }
}
func threeSumSmaller(nums []int, target int) int {
    sort.Ints(nums)
    n, ans := len(nums), 0
    for i := 0; i < n-2; i++ {
        l, r := i+1, n-1
        for l < r {
            sum := nums[i] + nums[l] + nums[r]
            if sum < target { ans += r - l; l++ } else { r-- }
        }
    }
    return ans
}
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size(), ans = 0;
        for (int i = 0; i < n - 2; i++) {
            int l = i + 1, r = n - 1;
            while (l < r) {
                int sum = nums[i] + nums[l] + nums[r];
                if (sum < target) { ans += r - l; l++; }
                else r--;
            }
        }
        return ans;
    }
};
class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        ans = 0
        for i in range(n - 2):
            l, r = i + 1, n - 1
            while l < r:
                s = nums[i] + nums[l] + nums[r]
                if s < target:
                    ans += r - l
                    l += 1
                else:
                    r -= 1
        return ans
var threeSumSmaller = function(nums, target) {
  nums.sort((a, b) => a - b);
  let ans = 0;
  for (let i = 0; i < nums.length - 2; i++) {
    let l = i + 1, r = nums.length - 1;
    while (l < r) {
      const sum = nums[i] + nums[l] + nums[r];
      if (sum < target) { ans += r - l; l++; }
      else r--;
    }
  }
  return ans;
};

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