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<1>题目描述
找一个二维数组中的鞍点，即该位置上的元素在该行上最大、在该列上最小。也可能没有鞍点
1 2 3 4 5 6 7 8 9 //3为该二维数组中鞍点
<2>思路分析
* 先找到第一行中最大的那个数a
* 将a与所在的列上其他数字进行比较，如果是最小，则输出a
* 如果a不是所在列最小的数，则找第二行最小的那个数b，重复第二步,直到找到鞍点
* 如果每一行中最大数都不是鞍点，则数组无鞍点
<3>代码实现
#include <stdio.h> #define ROW 3 //宏定义行 #define COL 3 //宏定义列 int main() { int
a[ROW][COL] = { {2,2,1},{4,5,6},{1,7,9} }; int k = 0; int j = 0; int i = 0; int
m = 0; for (i=0;i<ROW;i++) { int max = a[i][j]; //找第一行中最大的数字a for (j = 0; j <
COL; j++) { if (a[i][j] > max) { max = a[i][j]; //第二层循环走完以后，该max一定是该行中最大的那个数字 m
= j; //记录最大数字所在的列 } } j = 0; //判断a是否为所在列中最小的数 for (k = 0; k < ROW; k++) { if
(max > a[k][m]) { break; } } if (k == ROW) //表明a是该列中最小的数 { printf("鞍点为a[%d][%d]
= %d", i, m, max); break; } else k = 0; } if (i == ROW) printf("该二维数组没有鞍点\n");
return 0; }
<4>结果展示
