搜索二维矩阵

题目描述

给你一个满足下述两条属性的 m x n 整数矩阵：

• 每行中的整数从左到右按非严格递增顺序排列。
• 每行的第一个整数大于前一行的最后一个整数。 给你一个整数 target，如果 target 在矩阵中，返回 true；否则，返回 false。

示例 1：
输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
输出：true

示例 2：
输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
输出：false

思路&js代码

1、两次二分查找

var searchMatrix = function(matrix, target) {
    const rowIndex = binarySearchFirstColumn(matrix, target);
    if (rowIndex < 0) {
        return false;
    }
    return binarySearchRow(matrix[rowIndex], target);
};
 
const binarySearchFirstColumn = (matrix, target) => {
    let low = -1, high = matrix.length - 1;
    while (low < high) {
        const mid = Math.floor((high - low + 1) / 2) + low;
        if (matrix[mid][0] <= target) {
            low = mid;
        } else {
            high = mid - 1;
        }
    }
    return low;
}
 
const binarySearchRow = (row, target) => {
    let low = 0, high = row.length - 1;
    while (low <= high) {
        const mid = Math.floor((high - low) / 2) + low;
        if (row[mid] == target) {
            return true;
        } else if (row[mid] > target) {
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return false;
}
 

2、一次二分查找

var searchMatrix = function(matrix, target) {
    const m = matrix.length, n = matrix[0].length;
    let low = 0, high = m * n - 1;
    while (low <= high) {
        const mid = Math.floor((high - low) / 2) + low;
        const x = matrix[Math.floor(mid / n)][mid % n];
        if (x < target) {
            low = mid + 1;
        } else if (x > target) {
            high = mid - 1;
        } else {
            return true;
        }
    }
    return false;
};