leetcode_Range Sum Query 2D - Immutable

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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(1, 2, 2, 4) -> 12

Note:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

思路：

由题意可知，题目要求的是某矩形区域的元素之和，直接做这题可能难道不觉着难，但还不是很好下手的，但做过Range Sum Query - Immutable后感觉这题就比较显然了，分别快速求出该区域每一行的包括部分后相加即可求得题目要求的和。

代码：

public class NumArray {    int sumArr[]=null;    public NumArray(int[] nums) {        if(nums!=null&&nums.length!=0)        {             sumArr=nums;             for(int i=1;i<nums.length;i++)                sumArr[i]=sumArr[i-1]+nums[i];        }           }    public int sumRange(int i, int j) {        if(sumArr==null||sumArr.length==0)            return 0;        if(i==0)            return sumArr[j];        else            return sumArr[j]-sumArr[i-1];    }}// Your NumArray object will be instantiated and called as such:// NumArray numArray = new NumArray(nums);// numArray.sumRange(0, 1);// numArray.sumRange(1, 2);

当然了，还有更绝的方法，直接上代码了

private int[][] dp;public NumMatrix(int[][] matrix) {    if(   matrix           == null       || matrix.length    == 0       || matrix[0].length == 0   ){        return;       }    int m = matrix.length;    int n = matrix[0].length;    dp = new int[m + 1][n + 1];    for(int i = 1; i <= m; i++){        for(int j = 1; j <= n; j++){            dp[i][j] = dp[i - 1][j] + dp[i][j - 1] -dp[i - 1][j - 1] + matrix[i - 1][j - 1] ;        }    }}public int sumRegion(int row1, int col1, int row2, int col2) {    int iMin = Math.min(row1, row2);    int iMax = Math.max(row1, row2);    int jMin = Math.min(col1, col2);    int jMax = Math.max(col1, col2);    return dp[iMax + 1][jMax + 1] - dp[iMax + 1][jMin] - dp[iMin][jMax + 1] + dp[iMin][jMin];    }