leetcode:Search a 2D Matrix（数组，二分查找)

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：题意为在一个mxn矩阵中查找目标值。可以先通过二分法确定目标值target可能出现的行，然后再用一次二分法确定目标值target在行中的可能位置。

时间复杂度为O(logn+logm)

code如下：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int left=0;
        int right=matrix.size()-1;
        if(left != right){
            while(left <= right){
                int mid=left + (right-left)/2;
                if(matrix[mid][0]<target){
                    left=mid+1;
                }
                else if(matrix[mid][0]>target){
                    right=mid-1;
                }
                else {
                    return true;
                }
            }
        }
        if(right==-1){
            return false;
        }
        else{
            int row=right;
            int left=0;
            int right=matrix[row].size()-1;
            while(left<=right){
                int mid=left + (right-left)/2;
                if(matrix[row][mid]<target){
                    left=mid+1;
                }
                else if(matrix[row][mid]>target){
                    right=mid-1;
                }
                else {
                    return true;
                }
            }
            return false;
        }
    }
};

其他思路：

从左下角元素开始遍历，每次遍历中若与目标值target相等则返回true；若小于则列向右移动；若大于则行向下移动。时间复杂度O(logn+logm)

code如下：

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