题目:
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.
Example 1:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 3 Output: true
Example 2:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 13 Output: false
代码:
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int h = matrix.size();
if (h == 0)return false;
int w = matrix[0].size();
if (w == 0)return false;
if (target > matrix[h - 1][w - 1] || target < matrix[0][0])return false;
int bh = 0, bw = 0, eh = h - 1, ew = w - 1;
while (eh - bh > 1 || ew - bw > 1) {
int midh = bh + (eh - bh) / 2, midw = bw + (ew - bw) / 2;
if (matrix[midh][midw] == target)return true;
else if (matrix[midh][midw] > target) {
eh = midh;
ew = midw;
}
else {
bh = midh;
bw = midw;
}
}
for (int j = bw; j < w; j++) {
if (matrix[bh][j] == target)return true;
}
for (int j = 0; j <= ew; j++) {
if (matrix[eh][j] == target)return true;
}
return false;
}
};
想法:
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