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號，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;} };

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轉(zhuǎn)載于:https://www.cnblogs.com/qionglouyuyu/p/4854625.html

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