74. Search a 2D Matrix

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

链接： http://leetcode.com/problems/search-a-2d-matrix/

5/28/2017

1ms, 5%

注意的问题：

第11行，因为一开始的二分查找只是比较每行第一个元素，所以碰到比target小的时候，start = mid，而比target大的时候, end = mid - 1

第17行是为了让matrix[end][0]与target判断。或者可以删掉第17行，把第18行改成if (matrix[end][0] <= target)

第2个循环里就是正常的二分查找了，left = mid + 1, right = mid - 1的正常情况即可。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        int start = 0, end = matrix.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target) return true;
            else if (matrix[mid][0] < target) {
                start = mid;
            } else {
                end = mid - 1;
            }
        }
        int row;
        if (matrix[end][0] == target || matrix[start][0] == target) return true;
        if (matrix[end][0] < target) {
            row = end;
        } else {
            row = start;
        }
        int left = 0, right = matrix[0].length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (matrix[row][mid] == target) return true;
            else if (matrix[row][mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        if (matrix[row][left] == target || matrix[row][right] == target) return true;
        return false;
    }
}

别人的答案

把二维看作是一维二分查找，在整个0~m*n-1里面找。一次循环搞定

https://discuss.leetcode.com/topic/3227/don-t-treat-it-as-a-2d-matrix-just-treat-it-as-a-sorted-list

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int n = matrix.size();
        int m = matrix[0].size();
        int l = 0, r = m * n - 1;
        while (l != r){
            int mid = (l + r - 1) >> 1;
            if (matrix[mid / m][mid % m] < target)
                l = mid + 1;
            else 
                r = mid;
        }
        return matrix[r / m][r % m] == target;
    }
};

更多讨论

https://discuss.leetcode.com/category/82/search-a-2d-matrix