LeetCode304. Range Sum Query 2D - Immutable
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题目链接:
https://leetcode.com/problems/range-sum-query-2d-immutable/
题目描述:
求矩阵数组中由左上角点(row1,col1)和右下角点(row2,col2)形成的矩形中所有数字之和。
题目分析:
动态规划的思想。
sums[i][j]=sums[i-1][j]+sums[i][j-1]-sums[i-1][j-1]+matrix[i][j];
代码:
class NumMatrix {public: NumMatrix(vector<vector<int>> &matrix) { int row=matrix.size(); if(row==0){ return; } int col=matrix[0].size(); sums=vector<vector<int>>(row,vector<int>(col,0)); sums[0][0]=matrix[0][0]; for(int i=1;i<row;i++){ sums[i][0]=sums[i-1][0]+matrix[i][0]; } for(int i=1;i<col;i++){ sums[0][i]=sums[0][i-1]+matrix[0][i]; } for(int i=1;i<row;i++){ for(int j=1;j<col;j++){ sums[i][j]=sums[i-1][j]+sums[i][j-1]-sums[i-1][j-1]+matrix[i][j]; } } } int sumRegion(int row1, int col1, int row2, int col2) { if(row1>row2 || col1>col2){ return 0; } if(row1==0 && col1==0){ return sums[row2][col2]; } else if(row1==0){ return sums[row2][col2]-sums[row2][col1-1]; } else if(col1==0){ return sums[row2][col2]-sums[row1-1][col2]; } else{ return sums[row2][col2]-sums[row2][col1-1]-sums[row1-1][col2]+sums[row1-1][col1-1]; } }private: vector<vector<int>> sums;};
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