LeetCode Range Addition II
原题链接在这里:https://leetcode.com/problems/range-addition-ii/description/
题目:
Given an m * n matrix M initialized with all 0's and several update operations.
Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
You need to count and return the number of maximum integers in the matrix after performing all the operations.
Example 1:
Input: m = 3, n = 3 operations = [[2,2],[3,3]] Output: 4 Explanation: Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]] After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]] After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]] So the maximum integer in M is 2, and there are four of it in M. So return 4.
Note:
- The range of m and n is [1,40000].
- The range of a is [1,m], and the range of b is [1,n].
- The range of operations size won't exceed 10,000.
题解:
找出最小的更新row, 和最小的更新column, 返回乘机. 记得这两个可能比对应的m, n大, 所以初始值时m,n.
Time Complexity: O(ops.length). Space: O(1).
AC Java:
1 class Solution { 2 public int maxCount(int m, int n, int[][] ops) { 3 if(ops == null || ops.length == 0){ 4 return m*n; 5 } 6 7 int minRow = m; 8 int minColumn = n; 9 for(int [] op : ops){ 10 minRow = Math.min(minRow, op[0]); 11 minColumn = Math.min(minColumn, op[1]); 12 } 13 return minRow*minColumn; 14 } 15 }