利用 %20 替换 空格

将字符串中的空格都替换为 %20 （ 时间复杂度为O（N）的解法 ）

void ReplaceBlankSpace(char* arr){if (arr){int count = 0;int lenth = strlen(arr);for (int i = 0; i < lenth;++i)if (arr[i] == ' ')count++;char*before = arr+lenth, *behind = arr+lenth+2*count;while (before != behind){if (*before != ' '){*behind-- = *before--;}                else{*behind-- = '0';*behind-- = '2';*behind-- = '%';before--;}}}}

 

合并有序数组（ 时间复杂为O（N），空间复杂度为O（1）的解法）

void ArrayMerage(vector<int>& Vector1, vector<int>& Vector2){int index_1_before = Vector1.size() - 1, index_2 = Vector2.size() - 1;int sum = Vector1.size() + Vector2.size();int index_1_behind = sum - 1;Vector1.resize(sum);while (index_2 >= 0 && index_1_before >= 0)Vector1[index_1_behind--] = Vector1[index_1_before] > Vector2[index_2] ? \Vector1[index_1_before--] : Vector2[index_2--];while (index_2 >= 0)Vector1[index_1_behind--] = Vector2[index_2--];}