leetcode-1 Two Sum
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大多数答案都是使用C++和java的容器和hashmap解决的,O(n) runtime和O(n) space,网上也搜不到用C的解答
使用暴力破解法(O(n^2) runtime)提示time limited exceed
使用下面的方法,貌似遍历的次数少一些,可以通过oj
int res[2] = {0};int findLeft(int back[],int n,int left){ int i; for(i = 0; i < n; i++){ if(back[i] == left) return i; } return -1; }int *twoSum(int numbers[], int n, int target) { int i,j; int back[10240] = {0}; for(i = 0; i < n; i++){ int left = target - numbers[i]; int tmp = findLeft(back,i,left); if(tmp >= 0){ res[0] = tmp+1; res[1] = i + 1; return res; }else{ back[i] = numbers[i]; } }}
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