leetcode——74——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
.
class Solution {public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int i = 0; int j = matrix.size()-1; int mid = 0; //搜寻第一列 while(i <= j){ mid = (i + j)/2; if(matrix[mid][0] == target){ return true; }else if(matrix[mid][0] < target){ i = mid + 1; }else{ j = mid - 1; } }
<code class="java keyword"> if</code><code class="java plain">(m[mid][</code><code class="java value">0</code><code class="java plain">] > target){</code><div class="line number20 index19 alt1"><code class="java spaces"> </code><code class="java keyword">if</code><code class="java plain">(mid == </code><code class="java value">0</code><code class="java plain">){</code></div><div class="line number21 index20 alt2"><code class="java spaces"> </code><code class="java keyword">return</code> <code class="java keyword">false</code><code class="java plain">;</code></div><div class="line number22 index21 alt1"><code class="java spaces"> </code><code class="java plain">}</code></div><div class="line number23 index22 alt2"><code class="java spaces"> </code><code class="java plain">mid--;</code><code class="java comments">//mid-1</code></div><div class="line number24 index23 alt1"><code class="java spaces"> </code><code class="java plain">}</code></div> //搜寻mid行 i = 0; j = matrix[0].size() -1; int k = mid; //搜寻到了返回true while(i <= j){ mid = (i + j)/2; if(matrix[k][mid] == target){ return true; }else if(matrix[k][mid] < target){ i = mid + 1; }else{ j = mid - 1; } } //没有搜寻到,返回false return false; }};
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