LeetCode Search a 2D Matrix

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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.

思路：二分，或者从右上角，因为有序，所以当右上角的数小于目标的话就再往下一行找，依次类推。

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

        int cur = right;
        left = 0, right = matrix[0].size() - 1;
        while (left <= right) {
            int mid = left + right >> 1;
            if (matrix[cur][mid] == target) return true;
            else if (matrix[cur][mid] > target) right = mid - 1;
            else left = mid + 1;
        }

        return false;
    }
};

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

        return false;
    }
};

LeetCode Search a 2D Matrix

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原文地址：http://blog.csdn.net/u011345136/article/details/44888691