96 lines
2.7 KiB
Java
96 lines
2.7 KiB
Java
//编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
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//
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//
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// 每行中的整数从左到右按升序排列。
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// 每行的第一个整数大于前一行的最后一个整数。
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//
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
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//输出：true
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//
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//
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// 示例 2：
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//
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//
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//输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
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//输出：false
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//
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//
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//
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//
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// 提示：
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//
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//
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// m == matrix.length
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// n == matrix[i].length
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// 1 <= m, n <= 100
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// -104 <= matrix[i][j], target <= 104
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//
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// Related Topics 数组 二分查找 矩阵
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// 👍 478 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TwoArray;
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//74:搜索二维矩阵
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class SearchA2dMatrix {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new SearchA2dMatrix().new Solution();
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//true
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// TwoArray twoArray = new TwoArray("[[1,3,5,7],[10,11,16,20],[23,30,34,60]]",true);
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// System.out.println(solution.searchMatrix(twoArray.getArr(), 3));
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//true
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TwoArray twoArray = new TwoArray("[[1,3]]", true);
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System.out.println(solution.searchMatrix(twoArray.getArr(), 3));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public boolean searchMatrix(int[][] matrix, int target) {
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if (target < matrix[0][0] || target > matrix[matrix.length - 1][matrix[0].length - 1]) {
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return false;
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}
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int start = 0;
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int end = matrix.length - 1;
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while (start < end - 1) {
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int mid = (start + end) / 2;
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if (matrix[mid][0] == target) {
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return true;
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}
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if (matrix[mid][0] > target) {
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end = mid - 1;
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} else {
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start = mid;
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}
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}
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if (matrix[end][0] == target) {
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return true;
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}
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int xIndex = matrix[end][0] > target ? start : end;
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start = 0;
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end = matrix[0].length - 1;
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while (start < end) {
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int mid = (start + end) / 2;
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if (matrix[xIndex][mid] == target) {
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return true;
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}
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if (matrix[xIndex][mid] < target) {
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start = mid + 1;
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} else {
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end = mid;
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}
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}
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return matrix[xIndex][start] == target;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |