Leetcode 240. Search a 2D Matrix II [某二维数组从左往右升序，从上往下升序，查找某个值是否存在，存在则返回true，反之返回

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

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

Constraints:

m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-109 <= matix[i][j] <= 109
All the integers in each row are sorted in ascending order.
All the integers in each column are sorted in ascending order.
-109 <= target <= 109

package com.array;

/**
- @Author you guess
- @Date 2021/1/10 21:28
- @Version 1.0
- @Desc */ public class Leetcode_240_Searcha2DMatrixII {
  
  /**
  - 2021/1/10 自己写的，方法1
  - 先确定那几行的范围，再二分查找每行
  - Runtime: 8 ms, faster than 24.15% of Java online submissions for Search a 2D Matrix II.
  - Memory Usage: 51.3 MB, less than 8.84% of Java online submissions for Search a 2D Matrix II.
  - @param matrix
  - @param target
  - @return */ public boolean searchMatrix(int[][] matrix, int target) { int rows = matrix.length; int columns = matrix[0].length;
    
    int topRow = 0; for (int i = rows - 1; i >= 0; i--) { if (matrix[i][columns - 1] < target) { topRow = i; break; } }
    
    int bottomRow = rows - 1; for (int i = 0; i < rows; i++) { if (matrix[i][0] > target) { bottomRow = i - 1; break; } }
    
    for (int i = topRow; i <= bottomRow; i++) { if (binarySearch(matrix, target, i)) { return true; } } return false; }//searchMatrix
  public boolean binarySearch(int[][] matrix, int target, int row) {
```
 int head = 0;
 int tail = matrix[0].length - 1;

 while (head <= tail) {
     int mid = (head + tail) / 2;
     if (matrix[row][mid] == target) {
         return true;
     } else if (matrix[row][mid] < target) {
         head = mid + 1;
     } else {
         tail = mid - 1;
     }
 }
 return false;
```
  }
  
  /**
  - 2020-01-10，方法2
  - 参考【以右上角元素为基准】leetcode.com/problems/se…
  - 以左下角做基准，tar比当前元素大则往右找，tar比当前元素小则往上找！
  - Runtime: 4 ms, faster than 99.94% of Java online submissions for Search a 2D Matrix II.
  - Memory Usage: 44.8 MB, less than 34.46% of Java online submissions for Search a 2D Matrix II.
  - 时间复杂度O(N+M)
  - 这个方法是共用的！！
  - 可解决LeetCode 74. Search a 2D Matrix 【Runtime: 0 ms, faster than 100.00%】
  - @param matrix
  - @param target
  - @return */ public boolean searchMatrix2(int[][] matrix, int target) { int row = matrix.length - 1; int col = 0;
    
    while (row >= 0 && col < matrix[0].length) { if (matrix[row][col] == target) { return true; } else if (matrix[row][col] > target) { row--;//target比它小，在上边 } else { col++;//target比它大，在右边 } } return false; }
  public static void main(String[] args) { Leetcode_240_Searcha2DMatrixII main = new Leetcode_240_Searcha2DMatrixII(); // System.out.println(main.searchMatrix(new int[][]{{1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}} // , 5));//true
// System.out.println(main.searchMatrix2(new int[][]{{1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}} // , 14));//true
```
    System.out.println(main.searchMatrix2(new int[][]{{1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}}
            , 20));//false
}
```
}

方法2，是共用的！可解决LeetCode 74. Search a 2D Matrix 【Runtime: 0 ms, faster than 100.00%】