剑指offer题答案笔记

/*思路：二维数组从上到下，从左到右递增的规律选取右上角的元素与target比较target
    
     a[row][col],target在元素所在列的下边  row++target==a[row][col] 返回true当row > array[0].length -1 || col <0 没有找到*/public class Solution {    public boolean Find(int target, int [][] array) {        int row = 0;//行        int col = array[0].length -1;//列        while(row <= array[0].length -1 && col >=0){            if(target == array[row][col])                return true;            else if(target < array[row][col])                col--;            else                row++;                    }        return false;    }}
    

1.连续子数组的最大和

class Solution {public:    int FindGreatestSumOfSubArray(vector
    
      array) {        int size = array.size();        if(size == 0) return 0;        int maxSum = array[0];        int sum = 0;        for(int i= 0;i
     
      maxSum){                maxSum =sum;            }        }        return maxSum;    }};