算法学习（四）之 “二分查找”

1. 二分法

• 二分法模板代码：

public class BasicBinarySearch {
     
	public static void main(String[] args) {
     
		int[] num = {
     1,4,7,9,10,14,16,20,56,89};
		System.out.println(new BasicBinarySearch().getIndex(num, 5));
	}
	public int getIndex(int[] num, int target) {
     
		if (num == null || num.length == 0) {
     
			return -1;
		}
		int start = 0;
		int end = num.length - 1;
		int mid;
		while (start + 1 < end) {
     
			mid = start + (end - start) / 2;
			if (num[mid] == target) {
     
				return mid;
			} else if(num[mid] > target) {
     
				end = mid;
			} else {
     
				start = mid;
			}
		}
		if (num[start] == target) {
     
			return start;
		}
		if (num[end] == target) {
     
			return end;
		}
		return -1;
	}
}

2. 在旋转有序的数组中搜索 点我
   题目：
   
   图示：

代码：

class Solution {
     
    public int search(int[] num, int target) {
     
		if (num == null || num.length == 0) {
     
            return -1;
        }
        int start = 0;
        int end = num.length - 1;
        int mid;
        while (start + 1 < end) {
     
            mid = start + (end - start) / 2;
            if (num[mid] == target) {
     
                return mid;
            }
            if (num[mid] > num[start]) {
     
                if (target <= num[mid] && num[start] <= target) {
     
                    end = mid;
                } else {
     
                    start = mid;
                }
            } else {
     
                if (target >= num[mid] && target <= num[end]) {
     
                    start = mid;
                } else {
     
                    end = mid;
                }
            }
        }
        if (num[start] == target) {
     
            return start;
        }
        if (num[end] == target) {
     
            return end;
        }
        return -1;
	}
}

3. 在旋转有序的数组中找最小
   题目：

假设按照升序排序的数组在预先未知的某个点上进行了旋转。例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] 。
请找出其中最小的元素。

代码：

class Solution {
     
   public int findMin(int[] nums) {
     
		if (nums == null || nums.length == 0) {
     
            return 0;
        }
        int start = 0;
        int end = nums.length - 1;
        int mid;
       while (start + 1 < end ) {
     
           mid = start + (end - start) / 2;
           if (nums[mid] >= nums[start]) {
     
               if (nums[end] <= nums[mid]) {
     
                   start = mid;
               } else {
     
                   end = mid;
               }
           } else {
     
                end = mid; 
           }
       }
       return Math.min(nums[start], nums[end]);
	}
}

4. 找峰值元素

题目：

图示：
代码：

class Solution {
     
    public int findPeakElement(int[] nums) {
     
        if (nums == null || nums.length == 0) {
     
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int mid;
        while(start + 1 < end) {
     
            mid = start + (end - start) / 2;
            if (nums[mid] < nums[mid - 1]) {
     
                end = mid;
            } else if (nums[mid] < nums[mid + 1]) {
     
                start = mid;
            } else {
     
                return mid;
            }
        }
        return nums[start] > nums[end] ? start : end;
    }
}

6. 切木头

代码：

public class WoodCut {
     
	public static void main(String[] args) {
     
		int[] L = {
     232,124,456};
		System.out.println(new WoodCut().woodCut(L, 7));
	}
	public int woodCut(int[] L, int k) {
     
		if (L == null || L.length == 0) {
     
			return 0;
		}
		int start = 1;
		int end = getMax(L);
		int mid;
		while (start + 1 < end) {
     
			mid = start + (end - start) / 2;
			int pieces = getPieces(L, mid);
			if (pieces >= k) {
     
				start = mid;
			} else  {
     
				end = mid;
			}
		}
		if (getPieces(L, end) >= k) {
     
			return end;
		}
		if (getPieces(L, start) >= k) {
     
			return start;
		}
		return 0;
	}
	
	public int getMax(int[] L) {
     
		int max = L[0];
		for (int i = 1; i < L.length; i++) {
     
			if (max < L[i]) {
     
				max = L[i];
			}
		}
		return max;
	}
	
	public int getPieces(int[] L, int woodLength) {
     
		int pieces = 0;
		for (int wood: L) {
     
			pieces += wood/woodLength;
		}
		return pieces;
	}
}