LeetCode Top 100 Liked Questions 308. Range Sum Query 2D - Mutable (Java版; Hard)
题目描述
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left
corner (row1, col1) and lower right corner (row2, col2).
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3),
which contains sum = 8.
Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10
Note:
The matrix is only modifiable by the update function.
You may assume the number of calls to update and sumRegion function is distributed evenly.
You may assume that row1 ≤ row2 and col1 ≤ col2.
第一次做; 没有使用线段树或者树状数组, 而是多次使用了一维前缀和
classNumMatrix{privateint[][] matrix;privateint[][] sum;publicNumMatrix(int[][] matrix){this.matrix = matrix;getSumArr(matrix);}publicvoidupdate(int row,int col,int val){int diff = val - matrix[row][col];//记得更新matrix[row][col], 要不多次修改同一个位置后, 还是用最原始的matrix计算diff的值, 这样就错了!
matrix[row][col]= val;//细节: j的起始值, 要时刻牢记sum[i][j]的含义for(int j=col+1; j<sum[0].length; j++){
sum[row][j]= sum[row][j]+ diff;}}publicintsumRegion(int row1,int col1,int row2,int col2){int res =0;for(int i=row1; i<=row2; i++){
res = res + sum[i][col2+1]- sum[i][col1];}return res;}privatevoidgetSumArr(int[][] matrix){//input checkif(matrix==null || matrix.length==0|| matrix[0]==null || matrix[0].length==0)return;int rows = matrix.length, cols = matrix[0].length;/*
sum[i][j]表示matrix索引为i那一行的前j个数的和
sum[i][j]=matrix[i][0]+matrix[i][1]+...+matrix[i][j-1]
sum[i][j]=sum[i][j-1] + matrix[i][j-1]
*/
sum =newint[rows][cols+1];for(int i=0; i<rows; i++){for(int j=1; j<=cols; j++){
sum[i][j]= sum[i][j-1]+ matrix[i][j-1];}}}}/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix obj = new NumMatrix(matrix);
* obj.update(row,col,val);
* int param_2 = obj.sumRegion(row1,col1,row2,col2);
*/