LeetCode74 搜索二维矩阵

题目

解题

对二分查找的模板不熟悉的可以先去看：LeetCode704 二分查找。

解题一：一次二分查找

// javascript
var searchMatrix = function(matrix, target) {
    if (matrix === null || matrix.length === 0 || matrix[0].length === 0) return false;
    const m = matrix.length, n = matrix[0].length;
    let left = 0, right = m * n - 1;
    while (left <= right) {
        const mid = left + ((right - left) >> 1);
        const midVal = matrix[Math.floor(mid / n)][mid % n]; // 找到在数组中的位置
        if (midVal > target) right = mid - 1;
        else if (midVal < target) left = mid + 1;
        else return true;
    }
    return false;
};

解题二：两次二分查找

// javascript
var searchMatrix = function(matrix, target) {
    const rowIndex = binarySearchFirstColumn(matrix, target);
    if (rowIndex < 0) return false; // target < matrix[0][0]
    return binarySearchRow(matrix[rowIndex], target);
};

// 在第一列，找到最后一个小于等于 target 的元素所在行
const binarySearchFirstColumn = (matrix, target) => {
    let left = -1, right = matrix.length - 1;
    while (left < right) {
        const mid = left + ((right - left + 1) >> 1);
        if (matrix[mid][0] <= target) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
}

const binarySearchRow = (row, target) => {
    let left = 0, right = row.length - 1;
    while (left <= right) {
        const mid = left + ((right - left) >> 1);
        if (row[mid] > target) right = mid - 1;
        else if (row[mid] < target) left = mid + 1;
        else return true;
    }
    return false;
}

两种方法殊途同归，都利用了二分查找，在二维矩阵上寻找目标值。值得注意的是，若二维数组中的一维数组的元素个数不一，方法一将会失效，而方法二则能正确处理。

解题三：指针法

// javascript
var searchMatrix = function(matrix, target) {
    const m = matrix.length, n = matrix[0].length;
    let i = m - 1, j = 0;
    while (i >= 0 && j < n) {
        if (matrix[i][j] > target) i--;
        else if (matrix[i][j] < target) j++;
        else return true;
    }
    return false; 
};