leetcode 3Sum C++

　　荒废好久没更新了，时间过得很快，转眼就2017年了，经历了苦闷的科研阶段，发了论文顺利毕业；也经过三地辗转奔波来去的找工作，最终还是犹犹豫豫选择了自己知道以后可能会后悔的，果然就后悔了。所以还是应该选择自己喜欢的，当然也许人总是会这样，总觉得没走的另一条路也许走起来更畅通。

　　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.

　　解题思路：本题的思路与2 sum类似，首先对数组进行排序，然后依次把每位数字当做target，用类似于处理2 sum的思路进行两边扫描向中间逼近的处理。只是在此题中要注意重复的问题。

　　左数从i+1开始，可以理解为处理的sum[i]作为三个数中的最左值，从此数的右边第一位和数据的最后一位向中间靠拢寻找符合的数字组合。如果把sum[i]作为中间的值来处理则存在找出重复数组的风险，如下：

input：-1、-1、0、1、2output：[-1,-1,2] [0,-1,1] [1,-1,0]....//可以看到如果允许跨过target本身去寻找，就存在重复的风险

int left = i + 1;
int right = nums.size() - 1;

因为数组已经是排过序的，因此检查如下语句，避免作为左值的数重复。

if(i > 0 && nums[i] == nums[i - 1]) //continue;

　　数组中可能存在不止一组数字的和与target相加的值等于零，因此两边继续向内扫描，如下两句则是避免中间数字和右边数字重复的问题。

while(left < right && nums[left] == nums[left + 1]) left++;
while(left < right && nums[right] == nums[right -1 ]) right--;

　　完整代码如下：

class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {vector<vector<int>> resl;int length = nums.size();if(length < 3)return resl;sort(nums.begin(), nums.end());for (int i = 0; i < nums.size() - 1; ++i){if(i > 0 && nums[i] == nums[i - 1])continue;int left = i + 1;int right = nums.size() - 1;while(left < right){vector<int> temp;if(nums[i] + nums[left] + nums[right] == 0){temp.push_back(nums[i]);temp.push_back(nums[left]);temp.push_back(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[i] + nums[left] + nums[right] < 0)++left;else --right;if(temp.size() > 0)resl.push_back(temp);}}return resl;}
};

转载于:https://www.cnblogs.com/echo-lsh/p/6496911.html

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