2017-09-09 LeetCode_015 3Sum

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15. 3Sum

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.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

Solution:

class Solution {
public:
    vector<vector<int> > threeSum(vector<int>& nums) {
        vector<vector<int> > ans;
        if (nums.size() == 0) return ans;
        for (int i = 0; i < nums.size()-1; i++)
            for (int j = 0; j < nums.size()-i-1; j++)
                if (nums[j] > nums[j+1]) {
                    int t = nums[j]; nums[j] = nums[j+1]; nums[j+1] = t;
                }
        for (int b = 1; b < nums.size()-1;) {
            int tb = nums[b];
            int a = 0, c = nums.size()-1;
            if (tb < 0) {
                while (tb == nums[b] && b < nums.size()-1) b++;
                b--;
            } else if (tb == 0) {
                if (b+1 < nums.size()-1 && nums[b+1] == 0) b++;
            }
            while (a < b && b < c) {
                if (nums[a]+nums[b]+nums[c] == 0) {
                    vector<int> temp;
                    temp.push_back(nums[a]); temp.push_back(nums[b]); temp.push_back(nums[c]);
                    ans.push_back(temp);
                    int ta = nums[a];
                    while (ta == nums[a] && a < b) a++;
                } else if (nums[a]+nums[b]+nums[c] < 0) {
                    int ta = nums[a];
                    while (ta == nums[a] && a < b) a++;
                } else {
                    int tc = nums[c];
                    while (tc == nums[c] && b < c) c--;
                }
            }
            if (tb < 0) b++;
            else {
                while (tb == nums[b] && b < nums.size()-1) b++;
            }
        }
        return ans;
    }
};

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