Problem: 888. Fair Candy Swap 公平的糖果交换
排序，然后根据大小，依次查找，可以提前退出循环
更优秀的答案只需要一次遍历即可，
Code
class Solution { public: vector<int> fairCandySwap(vector<int>& aliceSizes, vector<int>& bobSizes) { sort(aliceSizes.begin(), aliceSizes.end()); sort(bobSizes.begin(), bobSizes.end()); int sa = 0, sb = 0, a, b; for(int& i : aliceSizes) sa += i; for(int& i : bobSizes) sb += i; if(sa < sb) { for(int i = 0; i < aliceSizes.size(); i++) { a = aliceSizes[i]; for(int j = bobSizes.size()-1; j >=0 ; j--) { b = bobSizes[j]; if(a >= b) break; if(sa - a + b == sb - b + a) { return {a, b}; } } } } else { for(int j = 0; j < bobSizes.size(); j++) { b = bobSizes[j]; for(int i = aliceSizes.size()-1; i >=0 ; i--) { a = aliceSizes[i]; if(a <= b) break; if(sa - a + b == sb - b + a) { return {a, b}; } } } } return {}; } };其他人更优秀的答案
class Solution { public: vector<int> fairCandySwap(vector<int>& aliceSizes, vector<int>& bobSizes) { int suma = 0, sumb = 0, lena = aliceSizes.size(), lenb = bobSizes.size(); for(int i = 0; i < lena; i++) { suma += aliceSizes[i]; } for(int j = 0; j < lenb; j++) { sumb += bobSizes[j]; } int num = (suma - sumb) / 2; sort(aliceSizes.begin(), aliceSizes.end()); sort(bobSizes.begin(), bobSizes.end()); vector<int> ans(2); for(int i = 0, j = 0; i < lena, j < lenb; i += 0, j += 0) { int n = aliceSizes[i] - bobSizes[j]; if(n == num) { ans[0] = aliceSizes[i]; ans[1] = bobSizes[j]; break; } else if(n > num) { j++; } else if(n < num) { i++; } } return ans; } };
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