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  • [LeetCode#74]Search a 2D Matrix

    The problem:

    Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

    • Integers in each row are sorted from left to right.
    • The first integer of each row is greater than the last integer of the previous row.

    For example,

    Consider the following matrix:

    [
      [1,   3,  5,  7],
      [10, 11, 16, 20],
      [23, 30, 34, 50]
    ]
    

    Given target = 3, return true.

    Solution1:

    public class Solution {
        public boolean searchMatrix(int[][] matrix, int target) {
            if (matrix == null || matrix.length == 0) //sterotype checking
                return false;
            int low = 0;
            int high = matrix.length - 1;
            int mid = -1;
            int row_index = -1;
            while (low <= high) { //search the row that may contain target
                mid = (low + high) / 2;
                if (matrix[mid][0] == target) {
                    return true;
                } else if (matrix[mid][0] < target) {
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
            /*a little skill: when exit while loop
            1. array[low] is the element just bigger than target;
            2. array[high] is th element just smaller than target.
            
            if (matrix[mid][0] > target) // the row that may contain target (mid + 1 is the possible to insert)
                row_index = mid - 1;
            else 
                row_index = mid; 
            */
            row_index = high;
            if (row_index < 0) //the target is even smaller than the smallest element in the matrix(first element)
                return false;
            low = 0;
            high = matrix[0].length - 1;
            mid = -1;
            while (low <= high) { //serach the target in the possible row
                mid = (low + high) / 2;
                if (matrix[row_index][mid] == target) {
                    return true;
                } else if (matrix[row_index][mid] < target) {
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
            return false;
        }
    }

    Solution2:

    public class Solution {
        public boolean searchMatrix(int[][] matrix, int target) {
            int row_count = matrix.length;
            int column_count = matrix[0].length;
            if (matrix == null || matrix[0][0] > target || matrix[row_count-1][column_count-1] < target)
                return false;
            int i = 0;
            int j = column_count - 1;
            while (i <= row_count-1 && j >= 0) {
                if (matrix[i][j] == target)
                    return true;
                else if (matrix[i][j] < target)
                    i++;
                else
                    j--;
            }
            return false;
        }
    }
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  • 原文地址:https://www.cnblogs.com/airwindow/p/4302505.html
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