Search a 2D Matrix ——Leetcode系列（十八）

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：首先對矩陣第一列進行二分查找，定位爲行下標index，滿足matrix[index][0] < target && matrix[index+1][0] > target

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int start = 0;
        int end = matrix.length - 1;//指向最後一個元素
        int pivot = 0;
        if(end == -1){
            return false;
        }
        while(start <= end){//是小於等於
            pivot = (start + end) / 2;
            if(target < matrix[pivot][0]){
                end = pivot-1;//減少1
            }else if(target > matrix[pivot][0]){
                start = pivot+1;//增加1
            }else{
                return true;
            }
        }
        int index;
        if(target < matrix[pivot][0]){
            index = pivot - 1;
        }else {
            index = pivot;
        }
        if(index < 0){
            return false;
        }
        start = 0;
        end = matrix[index].length - 1;
        while(start <= end){
            pivot = (start + end) / 2;
            if(target < matrix[index][pivot]){
                end = pivot -1;
            }else if(target > matrix[index][pivot]){
                start = pivot+1;
            }else{
                return true;
            }
        }
        return false;
    }
}