1. Topic
(The length of each one-dimensional array of same), each row from left to right in order of ascending sort, to sort each column in a two-dimensional array in order of increasing from top to bottom. A complete function, enter such a two-dimensional array and an integer, it is determined whether the array contains the integer.
2. Answers
/*
* 思路1：遍历每一行，二分查找是否属于这一行
*/
public static boolean find01(int[][] arr, int target) {
// 不在数组范围内直接返回false
if(target<arr[0][0] || target>arr[arr.length-1][arr[0].length-1])
return false;
for(int i=0; i<arr.length; i++) {
// 不在此行范围内继续下一行
if(target<arr[i][0]||target>arr[i][arr[i].length-1])
continue;
int left = 0;
int right = arr[i].length-1;
int mid = (arr[i].length)/2 ;
while(left <= right) {
if(target==arr[i][mid])
return true;
if(target>arr[i][mid]) {
left = mid + 1;
} else {
right = mid - 1;
}
mid = (left + right)/2;
}
}
return false;
}
/*
* 思路2：以对角线上的元素为原点将二维数组分成4个象限，从左上角向右下角移动原点，
* 若target在某两个原点之间，则可以去掉二四象限，从一三象限中寻找
* 也可不用从左上角向右下角移动原点，采用二分查找的思想，从中间开始
* 实现起来比较麻烦：弃
*/
/*
* 思路3：从右上角或者从左下角开始查找
* 以左下角为例：左下角元素是一列中的最大的元素,一行中最小的元素，当target<这个数则上移一行，否则右移一行
*/
public static boolean find03(int[][] arr, int target) {
// 不在数组范围内直接返回false
if(target<arr[0][0] || target>arr[arr.length-1][arr[0].length-1])
return false;
int index1 = arr.length-1;
int index2 = 0;
while(index1>=0&&index2<arr[0].length) {
int point = arr[index1][index2];
if(target == point)
return true;
if(target>point) {
index2++;
} else {
index1--;
}
}
return false;
}
Welcome to correct me, if any better ideas, welcome to discuss ~~
