74. Search a 2D Matrix (Graph; Divide-and-Conquer)

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. 所以用二分法
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.

思路：先按行二分搜索得到行号，再按列二分搜索

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

转载于:https://www.cnblogs.com/qionglouyuyu/p/4854625.html

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