LeetCode:Search a 2D Matrix

题目链接

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.                                                 本文地址

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分析：分两步，（1）先二分搜索的元素定位到行：当目标小于第一列某个元素时，向前面的行中去搜索；当目标大于第一列某个元素分两种情况 a、大于该元素所在行的最后一个元素时，往后面的行中去搜索，b、小于等于该元素所在行的最后一个元素，则可以定位到该元素所在的行。（2）在定位好的行中二分搜索

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int row = matrix.size();
        if(row == 0)return false;
        int col = matrix[0].size();
        int low = 0, high = row - 1;
        while(low < high)//注意这里没有=
        {//二分查找定位行
            int mid = (low + high) / 2;
            if(target < matrix[mid][0])
                high = mid - 1;
            else if(target > matrix[mid][0])
            {
                if(target > matrix[mid][col - 1])
                    low = mid + 1;
                else {low = mid; break;}
            }
            else return true;
        }
        int k = low; //已经把数据定位在了第row行
        low = 0;  high = col - 1;
        while(low <= high)
        {//行内二分查找
            int mid = (low + high) / 2;
            if(target < matrix[k][mid])
                high = mid - 1;
            else if(target > matrix[k][mid])
                low = mid + 1;
            else return true;
        }
        return false;
    }
};

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