Leetcode 240. Search a 2D Matrix II

再见小小ronnie

于 2017-01-22 03:23:55 发布

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分类专栏： Leetcode Binary Search

本文链接：https://blog.youkuaiyun.com/u011094114/article/details/54665819

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本文介绍了一种高效的矩阵搜索算法，该算法能在O(m+n)的时间复杂度内查找目标值。通过从矩阵右上角开始，根据目标值与当前值的比较结果决定向下一行或向左一列移动，逐步缩小搜索范围直至找到目标值或确认不存在。

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/**
 * Start from the top-right corner, check
 * if target > curr means we need to move to the next row b/c curr is the largest element in the row
 * if target < curr means we need to move to the previous column b/c curr is the samllest element in the column
 * each time we decrease the original problem by a row or column.
 * Time O(m+n).
 */ 
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) return false;
        
        // set curr as the top-right corner of the matrix
        int row = 0;
        int col = matrix[0].length-1;
        
        while (row < matrix.length && col >= 0) {
            if (matrix[row][col] == target) 
                return true;
            else if (matrix[row][col] < target)     // move to the next row
                row++;
            else                                    // move to the previous column
                col--;
        }
        
        return false;
    }
}