leetcode 074 —— 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：从左上角或者右下角的顶点开始搜索

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

第二种方法：最优路径搜索，也能通过，16ms，但是程序有待优化

class Solution {public:<span style="white-space:pre"></span>bool searchMatrix(vector<vector<int>>& matrix, int target) {<span style="white-space:pre"></span>vector<vector<int>> &a = matrix;<span style="white-space:pre"></span>int m = a.size();<span style="white-space:pre"></span>int n = a[0].size();<span style="white-space:pre"></span>int i = 0, j = 0;<span style="white-space:pre"></span>while (i<m&&j<n){<span style="white-space:pre"></span>if (a[i][j]>target) return false;<span style="white-space:pre"></span>if (a[i][j] == target) return true;<span style="white-space:pre"></span>if (i == m - 1){<span style="white-space:pre"></span>while (j < n){<span style="white-space:pre"></span>if (a[i][j] == target) return true;<span style="white-space:pre"></span>else if (a[i][j] < target) j++;<span style="white-space:pre"></span>else <span style="white-space:pre"></span>return false;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return false;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>if (j == n - 1){<span style="white-space:pre"></span>while (i < m){<span style="white-space:pre"></span>if (a[i][j] == target) return true;<span style="white-space:pre"></span>else if (a[i][j] < target) i++;<span style="white-space:pre"></span>else<span style="white-space:pre"></span>return false;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return false;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>if (a[i + 1][j] < target&&a[i][j + 1] < target){<span style="white-space:pre"></span>if (a[i + 1][j]>a[i][j + 1]) i++;<span style="white-space:pre"></span>else j++;<span style="white-space:pre"></span>continue;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>if (a[i + 1][j]>target&&a[i][j + 1]>target) return false;<span style="white-space:pre"></span>if (a[i + 1][j] == target || a[i][j + 1] == target) return true;<span style="white-space:pre"></span>if (a[i + 1][j]<target&&a[i][j + 1]>target){<span style="white-space:pre"></span>i++;<span style="white-space:pre"></span>continue;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>if (a[i + 1][j] > target&&a[i][j + 1] < target){<span style="white-space:pre"></span>j++;<span style="white-space:pre"></span>continue;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>}<span style="white-space:pre"></span>}}a;

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