需要在矩阵中查找元素。这个矩阵的排列如下：

每一行都是有序的。每一列都是有序的。

比如一个小矩阵。

10 30

20 80

现在，我们需要在一个这样N * M的矩阵中找到一个元素，并返回其位置。

思路

首先，这个题不太适合二分查找。因为并不能保证第二行的元素都一定比第一行的任意元素大。

所以应该是很难找到O(lgN)的算法。

每次都是取右上角的元素v与x（我们要查找的元素）进行比较较：

    如果发现v > x，由于v所在列肯定比x大。所以v所在列可以舍弃。

    如果发现v < x，由于v所在行肯定比x小。所以v所在行可能舍弃。

最后余下的，可能是一行，也可能是一列。总之可以利用二分查找来实现了。效率O(min(N, M)) + log(abs(M-N)).

解题

这里直接粘出代码。

int find(int **a, const int row, const int col, const int x, int *rpos, int *cpos) {
	//右上角为起始点.
	int from_row = 0, from_col = col - 1, v;
	int b, e, mid;
	*rpos = *cpos = -1;

	while (from_row < row && from_col >= 0) {
		v = a[from_row][from_col];
		if (v == x) {
			*rpos = from_row;
			*cpos = from_col;
			return 1;
		}
		from_row += x > v;
		from_col -= v > x;
	}
	//最后剩下一行
	if (from_row == (row - 1) && from_col != 0) {
		b = 0, e = from_col + 1;
		while (b < e) {
			mid = b + ((e-b)>>1);
			v = a[from_row][mid];
			if (x == v) {
				*rpos = from_row;
				*cpos = mid;
				return 1;
			} else if (v > x) e = mid;
			else b = mid + 1;
		}
		return 0;
	}
	//最后剩下一列
	if (from_col == 0 && from_row != (row - 1)) {
		b = from_row, e = row;
		while (b < e) {
			mid = b + ((e-b)>>1);
			v = a[mid][0];
			if (v == x) {
				*rpos = mid;
				*cpos = from_col;
				return 1;
			} else if (v > x) e = mid;
			else b = mid + 1;
		}
		return 0;
	}
	//最后只剩下一个点
	if (a[from_row][from_col] == x) {
		*rpos = from_row;
		*cpos = from_col;
		return 1;
	}
	return 0;
}

这里写一个测试程序,如果有错，会输出Error。

无限循环.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int _cmp(const void *a, const void *b) {
	int x = (*(const int *)a);
	int y = (*(const int *)b);
	return x - y;
}

int **create_matrix(int row, int col) {
	int *value = (int *) malloc(sizeof(int)*(row*col));
	int **a = (int **) malloc(sizeof(int *)*row);
	int i = 0, j = 0, iter = 0;
	const int numbers = row * col;

	for (i = 0; i < numbers ; ++i) {
		value[i] = rand() % 4096;
	}
	qsort(value, numbers, sizeof(int), _cmp);

	for (iter = i = 0; i < row; ++i) {
		a[i] = (int *) malloc(sizeof(int)*col);
		for (j = 0; j < col; ++j) {
			a[i][j] = value[iter++];
		}
	}

	free(value);
	return a;        
}

void destroy_matrix(int **a, const int row) {
	int i = 0;
	for (i = 0; i < row; ++i) {
		free(a[i]);
	}
	free(a);
}

void print_matrix(int **a, int row, int col) {
	int i, j;
	for (i = 0; i < row; ++i) {
		for (j = 0; j < col; ++j) {
			printf("%d ", a[i][j]);
		}
		printf("\n");
	}
}

int _binary_search(int *a, int from, int to, int x) {
	int *mid = NULL, *b = a + from, *e = a + to;
	while (b < e) {
		mid = b + ((e-b) >> 1);
		if (*mid == x) return mid - b;
		else if (*mid > x) e = mid;
		else b = mid + 1;
	}
	return -1;
}

int find(int **a, const int row, const int col, const int x, int *rpos, int *cpos) {
	//右上角为起始点.
	int from_row = 0, from_col = col - 1, v;
	int b, e, mid;
	*rpos = *cpos = -1;

	while (from_row < row && from_col >= 0) {
		v = a[from_row][from_col];
		if (v == x) {
			*rpos = from_row;
			*cpos = from_col;
			return 1;
		}
		from_row += x > v;
		from_col -= v > x;
	}
	//最后剩下一行
	if (from_row == (row - 1) && from_col != 0) {
		b = 0, e = from_col + 1;
		while (b < e) {
			mid = b + ((e-b)>>1);
			v = a[from_row][mid];
			if (x == v) {
				*rpos = from_row;
				*cpos = mid;
				return 1;
			} else if (v > x) e = mid;
			else b = mid + 1;
		}
		return 0;
	}
	//最后剩下一列
	if (from_col == 0 && from_row != (row - 1)) {
		b = from_row, e = row;
		while (b < e) {
			mid = b + ((e-b)>>1);
			v = a[mid][0];
			if (v == x) {
				*rpos = mid;
				*cpos = from_col;
				return 1;
			} else if (v > x) e = mid;
			else b = mid + 1;
		}
		return 0;
	}
	//最后只剩下一个点
	if (a[from_row][from_col] == x) {
		*rpos = from_row;
		*cpos = from_col;
		return 1;
	}
	return 0;
}

int main(void) {
	int row = 10, col = 10;
	int r = -1, c = -1;
	int x = rand() % row, y = rand() % col;
	int **a = NULL;
	while (true) {
		row = rand() % 100 + 1;
		col = rand() % 100 + 1;
		a = create_matrix(row, col);
		x = rand() % row, y = rand() % col;
		if (find(a, row, col, a[x][y] , &r, &c)) {
			if (a[x][y] != a[r][c]) {
				printf("Error\n");
			} 
		} else {
			printf("Error\n");
		}

		if (!find(a, row, col, a[0][0], &r, &c)) {
			printf("Error\n");
		}
		if (!find(a, row, col, a[row-1][0], &r, &c)) {
			printf("Error\n");
		}
		if (!find(a, row, col, a[0][col-1], &r, &c)) {
			printf("Error\n");
		}
		if (!find(a, row, col, a[row-1][col-1], &r, &c)) {
			printf("Error\n");
		}
		destroy_matrix(a, row);    
	}
	return 0;
}