LeetCode-74. Search a 2D Matrix

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.

首先确定在哪一行然后再进行二分查找。

package solutions._74;import jdk.nashorn.internal.ir.IfNode;class Solution {    private boolean BinarySearch(int[] arr, int target) {        int left = 0;        int right = arr.length - 1;        while (left <= right) {            int mid = left + (right - left) / 2;            if (arr[mid] > target) {                right = mid - 1;            } else if (arr[mid] < target) {                left = mid + 1;            } else {                return true;            }        }        return false;    }    public boolean searchMatrix(int[][] matrix, int target) {        if (matrix.length == 0 || matrix[0].length == 0) {            return false;        }        int i;        int row = -1;        for (i = 0; i < matrix.length - 1; i++) {            if (target >= matrix[i][0] && target < matrix[i + 1][0]) {                row = i;                break;            }        }        if (row == -1) {            row = i;        }        return BinarySearch(matrix[row], target);    }    public static void main(String[] args) {        Solution solution = new Solution();        int[][] matrix = new int[][]{ {1, 3, 5, 7}, {10, 11, 16, 20}, {23, 30, 34, 50}};        int target = 3;        System.out.println(solution.searchMatrix(matrix, target));    }}