[LeetCode]74. Search a 2D Matrix

Total Accepted: 73109 Total
Submissions: 218887 Difficulty: Medium

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,
Consider the following matrix:

[
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]

Given target =

3

,
return

true

.
思路：
先判断是不是小于最小值，大于最大值，如果不是
从右上角开始搜索，如果小于改行最右边的值，则在改行上进行二分查找；
否则，跳过此行，在下一行进行相同操作

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(target < matrix[0][0]||target > matrix[matrix.length-1][matrix[0].length-1]) return false;

boolean res = false;
boolean temp = false;
for (int i = 0;i < matrix.length; i++) {
if (target <= matrix[i][matrix[0].length-1]){
temp = searchRange(matrix[i], target);
if (temp == true) {
res = true;
break;
}
}

}
return res;
}
public boolean searchRange(int[] nums, int target) {
int low = 0;
int high = nums.length-1;
int mid = 0;
boolean r = false;
while(low <= high) {
mid = (low + high)/2;
if (nums[mid] == target) {

r = true;
break;
}
else if (nums[mid] < target) low = mid+1;
else high = mid -1;
}
//System.out.println(r);
return r;
}
}