[leetcode]240. Search a 2D Matrix II
2016-08-11 15:36
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public class Solution { public boolean searchMatrix(int[][] matrix, int target) { if(matrix.length==0 || matrix[0].length==0) return false; int i=0, j=matrix[0].length-1; while(i<matrix.length && j>=0) { int x = matrix[i][j]; if(target == x) return true; else if(target < x) --j; else ++i; } return false; } }
这道题的思路非常巧妙,本来我还是想用二分法做的,看了大神的代码觉得上述代码的方法更有效率。
思路是从数组的右上角开始搜索,如果matrix i j 比target小,则说明该行不可能有这个数,那么i++;如果matrix i j 比target大,则说明该列不可能有这个数,那么j–。
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