74. Search a 2D Matrix

https://leetcode.com/problems/search-a-2d-matrix/description/

使用 summary 里总结的 Binary Search 通用实现。两次 binary search：

• Left Column，寻找比 target 小的最大值rowIndex
• 在rowIndex所指row里，进行二分查找。

代码：

public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) {
        return false;
    }
    int rows = matrix.length;
    int cols = matrix[0].length;
    
    // not necessary. included in the logic below
    // if (matrix[0][0] > target || matrix[rows - 1][cols - 1] < target) {
    //     return false;
    // }
    
    int start = 0;
    int end = rows - 1;
    int mid, rowIndex;
    while (start <= end) {
        mid = start + (end - start) / 2;
        if (matrix[mid][0] == target) {
            return true;
        } else if(matrix[mid][0] < target) {
            start = mid + 1;
        } else {
            end = mid - 1;
        }
    }

    if (end == -1) {
        return false;
    }
    
    rowIndex = end;    
    start = 0;
    end = cols - 1;
    while(start <= end) {
        mid = start + (end - start) / 2;
        if (matrix[rowIndex][mid] == target) {
            return true;
        } else if (matrix[rowIndex][mid] < target) {
            start = mid + 1;
        } else {
            end = mid - 1;
        }
    }
    
    return false;
}

另一种思路是将该 matrix 理解为长度为 m*n 的 array，进行二分。index 需要转成 rowIndex，colIndex，来比较 item 与 target。

rowIndex = index / colsNum;
colIndex = index % colsNum;