Search a 2D matrix

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本文介绍了一种高效的矩阵搜索算法，该算法能在排序矩阵中快速查找指定数值。通过在第一列找到目标值的下界，然后进行二分搜索实现。具体步骤包括：1. 在第一列中寻找目标值的下界；2. 对目标行使用二分搜索查找目标值。

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

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

Analysis: Found the lower bound in the first column first and then do a binary search.

public class Solution {
    public int findLowerBound(int[] A, int target) {
        int low=0, high=A.length-1;
        while(low < high) {
            int mid = (low+high)/2+1;
            if(A[mid]==target) return mid;
            else if(A[mid]>target) high=mid-1;
            else if(A[mid]<target) low=mid;
        }
        return low;
    }
    
    public boolean binarySearch(int[] B, int target) {
        int low=0, high=B.length-1;
        while(low <= high) {
            int mid = (low+high)/2;
            if(B[mid]==target) return true;
            else if(B[mid]>target) high=mid-1;
            else if(B[mid]<target) low=mid+1;
        }
        return false;
    }
    
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length==0) return false;
        if(target<matrix[0][0]) return false;
        int row=matrix.length, col=matrix[0].length;
        int[] firstCol = new int[row];
        for(int i=0; i<row; i++) {
            firstCol[i] = matrix[i][0];
        }
        int lowerBound = findLowerBound(firstCol, target);
        int[] searchRow = Arrays.copyOf(matrix[lowerBound], col);
        return binarySearch(searchRow, target);
    }
}