LeetCode *** 74. 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。矩阵特性为每行元素从左到右递增排序，且每行首元素大于前一行末元素。通过分治策略实现高效查找。

题目：

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.

分析：

分治。

代码：

class Solution {
public:
    int myt;
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size()<1)return false;
        myt=target;
        return search(matrix,0,matrix.size()-1,0,matrix[0].size()-1);
    }
    
    bool search(vector<vector<int>>& matrix,int i1,int i2,int j1,int j2){
        if(matrix[i1][j1]>myt)return false;
        if(i1>i2||j1>j2)return false;
        if(i1==i2&&j1==j2){
            if(matrix[i1][j1]==myt)return true;
            else return false;
        }
        
        if(i1!=i2){
            int midi=(i1+i2)/2;
            if(matrix[midi][j2]<myt) return search(matrix,midi+1,i2,j1,j2);
            else return search(matrix,i1,midi,j1,j2);
        }
        
        else {
            int midj=(j1+j2)/2;
            if(matrix[i1][midj]<myt) return search(matrix,i1,i2,midj+1,j2);
            else return search(matrix,i1,i2,j1,midj);
        }
        
        
    }
};