303. 区域和检索 - 数组不可变(中等)
题目:
思路:
用前缀和记录开始到当前位置的元素和,这里前缀和数组长度n+1,presum[n+1]代表0-n的元素和
class NumArray { public: NumArray(vector<int>& nums) { int n=nums.size(); presum.resize(n+1); for(int i=1;i<=n;++i){ presum[i]=presum[i-1]+nums[i-1]; } } int sumRange(int left, int right) { return presum[right+1]-presum[left]; } vector<int> presum; }; /** * Your NumArray object will be instantiated and called as such: * NumArray* obj = new NumArray(nums); * int param_1 = obj->sumRange(left,right); */
304. 二维区域和检索 - 矩阵不可变(中等)
题目:
思路:
前缀和数组记录[0][0]->[i][j]的所有元素和。为了方便求值,在遍历时presum[i][j]=presum[i-1][j](原点到上边一点的矩形)+presum[i][j-1](原点到左边一点的矩形)-presum[i-1][j-1](原点到左上方的矩形)+matrix[i-1][j-1](加上当前点)
然后求值时presum[row2+1][col2+1](绿色)-presum[row2+1][col1](橙色)-presum[row1][col2+1](蓝色)+presum[row1][col1](粉色)
class NumMatrix { public: NumMatrix(vector<vector<int>>& matrix) { int m=matrix.size(); int n=matrix[0].size(); presum=vector<vector<int>>(m+1,vector<int>(n+1)); for(int i=1;i<=m;++i){ for(int j=1;j<=n;++j){ presum[i][j]=presum[i-1][j]+presum[i][j-1]-presum[i-1][j-1]+matrix[i-1][j-1]; } } } int sumRegion(int row1, int col1, int row2, int col2) { return presum[row2+1][col2+1]-presum[row2+1][col1]-presum[row1][col2+1]+presum[row1][col1]; } vector<vector<int>> presum; }; /** * Your NumMatrix object will be instantiated and called as such: * NumMatrix* obj = new NumMatrix(matrix); * int param_1 = obj->sumRegion(row1,col1,row2,col2); */
560. 和为K的子数组(中等)
题目:
思路:
用哈希表记录当前的前缀和
然后在累加前缀和时,从哈希表中查找相减等于k的值,并累加其次数
class Solution { public: int subarraySum(vector<int>& nums, int k) { unordered_map<int,int> m; int sum=0; //base case,当前缀和本身==k时,会查找到0 m[0]=1; int ret=0; for(int i=0;i<nums.size();++i){ sum+=nums[i]; //查找与sum相减为k的位置 int other=sum-k; if(m.count(other)>0){ ret+=m[other]; } //累加前缀和为sum的次数 m[sum]++; } return ret; } };