• 博客园logo
  • 会员
  • 众包
  • 新闻
  • 博问
  • 闪存
  • 赞助商
  • HarmonyOS
  • Chat2DB
    • 搜索
      所有博客
    • 搜索
      当前博客
  • 写随笔 我的博客 短消息 简洁模式
    用户头像
    我的博客 我的园子 账号设置 会员中心 简洁模式 ... 退出登录
    注册 登录
neverlandly
博客园    首页    新随笔    联系   管理    订阅  订阅

Leetcode: 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).
Range Sum Query 2D
Range Sum Query 2D
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
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
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.

 

 1 class NumMatrix {
 2 
 3     private int[][] dp;
 4 
 5     public NumMatrix(int[][] matrix) {
 6         if(   matrix           == null
 7            || matrix.length    == 0
 8            || matrix[0].length == 0   ){
 9             return;   
10         }
11 
12         int m = matrix.length;
13         int n = matrix[0].length;
14 
15         dp = new int[m + 1][n + 1];
16         for(int i = 1; i <= m; i++){
17             for(int j = 1; j <= n; j++){
18                 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] -dp[i - 1][j - 1] + matrix[i - 1][j - 1] ;
19             }
20         }
21     }
22 
23     public int sumRegion(int row1, int col1, int row2, int col2) {
24         return dp[row2 + 1][col2 + 1] - dp[row2 + 1][col1] - dp[row1][col2 + 1] + dp[row1][col1];    
25     }
26 }

 

posted @ 2015-12-29 09:25  neverlandly  阅读(323)  评论(0)    收藏  举报
刷新页面返回顶部
博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3