LeetCode Search a 2D Matrix

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rows, cols;
        if (!(rows = matrix.size()) || !(cols = matrix[0].size())) return false;
        int ridx, cur, first = -1, last = rows;
        while (first + 1 < last) {
            ridx = (first + last) / 2;
            cur = matrix[ridx][0];
            if (cur < target) {
                first = ridx;
            } else if (cur > target) {
                last = ridx;
            } else {
                break;
            }
        }
        if (first + 1 == last) {
            ridx = first;
        } else {
            return true;
        }
        if (ridx < 0 || ridx >= rows) return false;
        return binary_search(matrix[ridx].begin(), matrix[ridx].end(), target);
    }
};

先用二分找到合适范围的行索引，然后在该行上搜索，在整个范围上搜索的话也可以貌似复杂度更低一点，代码如下

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rows, cols;
        if (!(rows = matrix.size()) || !(cols = matrix[0].size())) return false;
        int first = 0, last = rows * cols - 1;
        while (first <= last) {
            int mi = (first + last) / 2;
            int mv = matrix[mi / cols][mi % cols];
            if (mv > target) {
                last = mi - 1;
            } else if (mv < target) {
                first= mi + 1;
            } else {
                return true;
            }
        }
        return false;
    }
};

二分搜索android sdk里有个代码感觉很不错

// 未找到，返回应该插入的索引位置的负值
private static int binarySearch(int[] a, int start, int len, int key) {
    int high = start + len, low = start - 1, guess;

    while (high - low > 1) {
        guess = (high + low) / 2;

        if (a[guess] < key)
            low = guess;
        else
            high = guess;
    }

    if (high == start + len)
        return ~(start + len);
    else if (a[high] == key)
        return high;
    else
        return ~high;
}