74. 搜索二维矩阵（算法200题-42）

一、二维数组转一维数组 [ [a, b] ] => [ c, d ]
一维数组的 index = row * 二维下标x + 二维下标y

一维数组转二维数组 matrix = [ [a, b] ]
二维下标 X = Math.floor(index / row)
二维下标 Y = Math.floor(index % row)

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
    if(matrix === null || !matrix.length) return false
    let row = matrix[0].length,
        col = matrix.length,
        l = 0,
        r = (col - 1) * row + row - 1
    while (l <= r) {
        const m = Math.floor(l + (r - l) / 2)
        const cur = matrix[Math.floor(m / row)][Math.floor(m % row)]
        if(cur === target) return true
        if(cur > target) {
            r = m - 1
        } else {
            l = m + 1
        }
    }
    return false
};