leetcode[Range Addition II]
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解法一:
class Solution { /*public int maxCount(int m, int n, int[][] ops) { int[][] matrix = new int[m][n]; int max = 0; for(int i = 0; i < ops.length; i++){ int a = ops[i][0]; int b = ops[i][1]; //System.out.println("a:" + a + " b:" + b); for(int j = 0; j < m && j < a; j++){ for(int k = 0; k < n && k < b; k++){ //System.out.println("i:" + i + " j:" + j + " k:" + k); matrix[j][k]++; max = Math.max(max, matrix[j][k]); } } } //统计有多少个max int count = 0; for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ //System.out.print(matrix[i][j] + " "); if(matrix[i][j] == max){ count++; } } //System.out.println(); } return count; }*///不能向上面那样一直去遍历矩阵,数量过大时会造成超时//operations = [[2,2],[3,3]],归根到底改变的还是某一行,某一列,最大的元素肯定是每次都被操作到了的//比如第一行第一列的元素,肯定每次都被操作到了,所以只要求ops中行和列的交集public int maxCount(int m, int n, int[][] ops) {int row = m;int col = n;for(int i = 0; i < ops.length; i++){row = Math.min(row, ops[i][0]);col = Math.min(col, ops[i][1]);}return row*col;}}
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