力扣 题目74- 搜索二维矩阵

题目

题解

先对第一列进行二分搜索 找到对应行

再对 对应行进行二分搜索

代码

#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        //1.二分搜索找行 如果存在直接返回
        int xnumber = matrix.size()-1;
        int ynumber = matrix[0].size()-1;
        int line = xnumber / 2;
        int left = 0;
        int right = xnumber;
        while (true)
        {
            line = (left + right) / 2 ;
            if (matrix[line][0] == target) {
                return 1;
            }
            if (left == right || right < left) {
                //在二分搜索中 最后的结果有可能是 matrix[left][0] > target;
                //我们要找到是比target小的 所以left=left - 1
                if (matrix[left][0] > target) {
                    left=left - 1;
                }
                //left < 0 说明target<matrix[0][0] 肯定没有这个数
                if (left < 0) {
                    return 0;
                }
                line = left;
                break;
            }
            else if(matrix[line][0] < target)
            {
                left = line + 1;
            }
            else if (matrix[line][0] > target)
            {
                right= line - 1;
            }
        }
        //2.找到行再进行二分搜索
        left = 0;
        right = ynumber;
        int center = (right + left) / 2;
        while (true)
        {
            center = (left + right) / 2;
            if (matrix[line][center] == target) {
                return 1;
            }
            if (left == right|| right< left) {
                return 0;
            }
            else if (matrix[line][center] < target)
            {
                left = center + 1;
            }
            else if (matrix[line][center] > target)
            {
                right = center - 1;
            }
        }
        return 0;
    }
};

int main() {
    Solution sol;

    vector<vector<int>> matrix = { {1} };
    bool result=sol.searchMatrix(matrix,0);
    cout << result << endl;
}