LeetCode - 598. Range Addition II

标签：tput   部分   class   inpu   logs   span   code   after   ber   

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.

模拟会空间复杂度会超，必须直接计算。计算重叠部分。

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length <=0 || ops[0].length <= 0)
            return m*n;
        int minX = Integer.MAX_VALUE, minY = Integer.MAX_VALUE;
        for (int i=0; i<ops.length; i++) {
            int p = ops[i][0], q = ops[i][1];
            if (minX > p)
                minX = p;
            if (minY > q)
                minY = q;
        }
        return minX * minY;
    }
}

LeetCode - 598. Range Addition II

标签：tput   部分   class   inpu   logs   span   code   after   ber   

原文地址：http://www.cnblogs.com/wxisme/p/7462482.html