找到矩阵区域最小数值
找到矩阵区域最小数值a[i][j]
满足条件
a[i][j]<a[i+1][j],
a[i][j]<a[i][j+1],
a[i][j]<a[i-1][j],
a[i][j]<a[i][j-1]
方法为：
1. 取矩阵中间Row，找到最小值，记录curRow，curCol，curMin
2. 将此最小值与四周值比较，如果有比curMin小的值，则更新curRow，curCol，curMin，重复第二步。如果不存在，则返回当前值，即为结果。
#include <cstdio>const int N = 5;const int M = 5;int a[N][M] = {1, 12, 3, 1, -23, 7, 9, 8, 5, 6, 4, 5, 6, -1, 77, 7, 0, 35, -2, -4, 6, 83, 1, 7, -6};int direct[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };void findLocalMin(int& resRow, int& resCol){int curRow = N / 2;int curCol = 0;int curMin = a[curRow][curCol];for(int i = 0; i < M; i++){if (a[curRow][i] < curMin){curMin = a[curRow][i];curCol = i;}}bool flag = false;while(true){flag = true;for (int i = 0; i < 4; i++){int tempRow = curRow + direct[i][0];int tempCol = curCol + direct[i][1];if (tempRow < N && tempCol < N && a[curRow][curCol] > a[tempRow][tempCol]){curMin = a[tempRow][tempCol];curRow = tempRow; curCol = tempCol;flag = false;break;}}if (flag){resRow = curRow; resCol = curCol;return;}}};int main(){int resRow, resCol;findLocalMin(resRow, resCol);printf("%d", a[resRow][resCol]);return 0;}