第十八周：[Leetcode]74. Search a 2D Matrix

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本文介绍了一种高效的算法，用于在一特殊性质的二维有序矩阵中查找特定值。该矩阵每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章详细解释了如何通过比较每行最后一个元素来缩小搜索范围，再利用二分搜索来定位目标值。

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.

已排序的矩阵，先对比每行最后一个值，若比目标值大，则目标值肯定在该行搜索，而后使用二分搜索。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() == 0 || matrix[0].size() == 0)
            return false;
        int i = 0,j = matrix[0].size() - 1;
        while(i < matrix.size()){
            if(matrix[i][j] == target)
                return true;
            if(matrix[i][j] < target)
                i++;
            else{
                int k = 0;
                j --;
                while(k <= j){
                    int mid = k + ((j - k) >> 1);
                    if(matrix[i][j] == target || matrix[i][k] == target || matrix[i][mid] == target)
                        return true;
                    if(matrix[i][mid] > target)
                        j = mid - 1;
                    else
                        k = mid + 1;
                }
                return false;
            }
        }
        return false;
    }
};