Search 2D matrix II

Search 2D matrix II

在矩阵中搜索目标数字的出现次数。

Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

    * Integers in each row are sorted from left to right.

    * Integers in each column are sorted from up to bottom.

    * No duplicate integers in each row or column. 无重复数字

Example

Consider the following matrix:

[

    [1, 3, 5, 7],

    [2, 4, 7, 8],

    [3, 5, 9, 10]

]

Given target = 3, return 2.

Solution:

public int searchMatrix(int[][] matrix, int target) {    if(matrix == null || matrix.length == 0) {        return 0;    }    int row = matrix.length - 1;    int col = matrix[0].length - 1;    int count = 0;    for(int i = 0; i <= row; i++) {        if(target >= matrix[i][0] && target <= matrix[i][col]) {            count += bs(matrix[i], target);        }    }    return count;}private int bs(int[] nums, int target) {    int start = 0;    int end = nums.length - 1;    int mid;    while(start + 1 < end) {        mid = start + ((end - start) >> 1);        if(target < nums[mid]) {            end = mid;        } else if(target > nums[mid]) {            start = mid;        } else {            return 1;        }    }    if(target == nums[start]) {        return 1;    } else if(target == nums[end]) {        return 1;    }  else {        return 0;                }}

 一次遍历：

    public int searchMatrix(int[][] matrix, int target) {        if (matrix == null || matrix.length == 0) {            return 0;        }        if (matrix[0] == null || matrix[0].length == 0) {            return 0;        }        int row = matrix.length - 1;        int col = 0;        int size = matrix[0].length - 1;        int count = 0;        while (row >= 0 && col <= size) {            if (matrix[row][col] == target) {                count++;                row--;            } else if (matrix[row][col] < target) {                col++;            } else if (matrix[row][col] > target) {                row--;            }        }        return count;    }

思路：

1. 搜索：对每一行进行bs，统计所有行 O(mlog(n))。每行没有重复，每行最后一个是最大值。 如果target比它大，则该行可略过。

2. 一次遍历。从左下角开始向右上前进，每行没有重复。同一列中，上一行比本行小。