leetcode 304. Range Sum Query 2D - Immutable
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).
![](https://img.laitimes.com/img/9ZDMuAjOiMmIsIjOiQnIsICNyMTNwgDMwIjMwETM2EDMy8CX0Vmbu4GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.jpg)
Example:
Given matrix = [
[, , , , ],
[, , , , ],
[, , , , ],
[, , , , ],
[, , , , ]
]
sumRegion(, , , ) ->
sumRegion(, , , ) ->
sumRegion(, , , ) ->
Note:
You may assume that the matrix does not change.
There are many calls to sumRegion function.
You may assume that row1 ≤ row2 and col1 ≤ col2.
解析:
本来自己在做的时候,直接就去遍历,发现超时,没有往动态规划这块去想,就是缺少一个明确的定义。动态规划的第一步就是要有一个明确的定义。
lc上discuss中优美的答案,首先定义一个比原矩阵大一圈的矩阵dp[m+1][n+1],便于后续边界条件的判断。
dp[i][j]的含义是在(i, j)之前,所有元素的和,不包括(i, j),那么整个矩阵的和就是dp[m+1][n+1]。
而且dp矩阵的初始化需要使用动态规划的思路。
如下图
显然
dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]
+ m[i-1][j-1];
后续的public int sumRegion(int row1, int col1, int row2, int col2)函数可以使用几个矩阵的运算来做
公式为
紫色面积 = dp[row2+1][col2+1] - dp[row1][col2+1]
- dp[row2+1][col1]
- + dp[row1][col1];
具体的代码实现如下
public class NumMatrix {
public static int[][] m;
public static int[][] dp;
public NumMatrix(int[][] matrix) {
if(matrix == null || matrix.length == || matrix[].length == ){
m = null;
}
else{
m = new int[matrix.length][matrix[].length];
for(int i = ; i < matrix.length; i++){
for(int j = ; j < matrix[].length; j++){
m[i][j] = matrix[i][j];
}
}
dp = new int[matrix.length+][matrix[].length+];
for(int i = ; i <= matrix.length; i++){
for(int j = ; j <= matrix[].length; j++){
dp[i][j] = dp[i-][j] + dp[i][j-] - dp[i-][j-] + m[i-][j-];
}
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
return dp[row2+][col2+] - dp[row1][col2+] - dp[row2+][col1] + dp[row1][col1];
}
}
// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);