LeetCode.303/304 Range Sum Query(2) - Immutable
作者：互联网
303.题目（求数组子序列元素之和）
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
Example:
Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
Note:
You may assume that the array does not change.
There are many calls to sumRange function.
303.分析
class NumArray {
// 思路：统计每个元素从index=0到当前元素的下表的元素之和
private int[] nums;
private int[] sums;
public NumArray(int[] nums) {
this.nums=nums;
if(nums!=null&&nums.length>0){
sums=new int[nums.length];
sums[0]=nums[0];
for(int i=1;i<nums.length;i++){
sums[i]=sums[i-1]+nums[i];
}
}
}
public int sumRange(int i, int j) {
if(nums==null||nums.length==0){
return 0;
}
if(i!=0){
return sums[j]-sums[i-1];
}
return sums[j];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
304.题目（求2维数组的元素之和）
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
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.
304.分析
class NumMatrix {
// 思路：按照给定的方位点求出对应的累计和，最后将起始点的累计和进行相减（分别针对每一行的数据）,同303的解法类似
private int[][] matrix;
private int[][] sums;
public NumMatrix(int[][] matrix) {
this.matrix=matrix;
if (matrix != null && matrix.length > 0 && matrix[0].length > 0) {
sums = new int[matrix.length][matrix[0].length];
}
initSumsArray(matrix,sums);
}
public int sumRegion(int row1, int col1, int row2, int col2) {
int res=0;
for(int i=row1;i<=row2;i++){
if(col1!=0){
res+=sums[i][col2]-sums[i][col1-1];
}else{
res+=sums[i][col2];
}
}
return res;
}
public void initSumsArray(int [][] matrix,int [][] sums){
if(matrix==null||matrix.length==0||matrix[0].length==0){
return ;
}
int rowLength=matrix.length;
int colLength=matrix[0].length;
for(int i=0;i<rowLength;i++){
for(int j=0;j<colLength;j++){
if(j==0){
// 初始化第一列
sums[i][j]=matrix[i][j];
}else{
sums[i][j]=sums[i][j-1]+matrix[i][j];
}
}
}
}
}
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix obj = new NumMatrix(matrix);
* int param_1 = obj.sumRegion(row1,col1,row2,col2);
*/
xiakexiaohu
发布了296 篇原创文章 · 获赞 59 · 访问量 34万+
他的留言板
关注
标签：matrix,nums,int,sums,Sum,304,LeetCode.303,length,sumRegion 来源： https://blog.csdn.net/xiakexiaohu/article/details/104119158