1 题目理解

输入：一个mxn的int数组matrix，这个数组每一行按照从小到大排序，并且下一行的第一个值大于上一行的最后一个值。一个int值target
返回：target在matrix中是否存在。存在返回true。

2 二分

我们可以把这个数组看做是一个有序数组。数组下表从0到m*n-1。

class Solution {public boolean searchMatrix(int[][] matrix, int target) {if(matrix==null || matrix.length==0 || matrix[0].length==0) return false;int l = 0;int r = matrix.length*matrix[0].length-1;int cols = matrix[0].length;while(l<=r){int m = l + ((r-l)>>1);int row = m/cols;int col = m%cols;if(matrix[row][col]==target) return true;if(matrix[row][col]>target){r = m - 1;}else{l = m + 1;}}return false;}
}

也可以先在第一列查找，找到最小的比target大的元素，所在的行l，然后在l-1行查找target是否存在。

class Solution {public boolean searchMatrix(int[][] matrix, int target) {if(matrix == null || matrix.length==0 || matrix[0].length==0) return false;int m = matrix.length;int n = matrix[0].length;//先在第一列找,l是nums[l]>target的最小值int l = 0, r = m-1;while(l<=r){int middle = l+((r-l)>>1);if(matrix[middle][0] == target) return true;if(matrix[middle][0]>target){r = middle-1;}else{l = middle+1;}}if(l>0){int line = l-1;//在line这一行查找l = 0;r = n-1;while(l<=r){int middle = l +((r-l)>>1);if(matrix[line][middle]==target) return true;if(matrix[line][middle]>target){r = middle-1;}else{l = middle+1;}}return false;}else{return false;}}
}

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1 题目理解

2 二分

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