标签：记录 osi 递归 man help style public str for
You are given a list of non-negative integers, a1, a2, ..., an, and a target, S.
Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol. Find out how many ways to assign symbols to make sum of integers equal to target S. Example 1: Input: nums is [1, 1, 1, 1, 1], S is 3. Output: 5 Explanation: -1+1+1+1+1 = 3 +1-1+1+1+1 = 3 +1+1-1+1+1 = 3 +1+1+1-1+1 = 3 +1+1+1+1-1 = 3 There are 5 ways to assign symbols to make the sum of nums be target 3. Note: The length of the given array is positive and will not exceed 20. The sum of elements in the given array will not exceed 1000. Your output answer is guaranteed to be fitted in a 32-bit integer.
思路: 先想到递归, 但是1要求是count, 2画图发现有重叠递归的部分, 可以使用记忆化搜索记录中间值, 所以想到动归
记忆化搜索常用分治法, 将回溯后的值存入map:
String encodeString = index + "->" + sum;
if (map.containsKey(encodeString)){
return map.get(encodeString);
}
public int findTargetSumWays(int[] nums, int S) {
if (nums == null || nums.length == 0){
return 0;
}
return helper(nums, 0, 0, S, new HashMap<>());
}
private int helper(int[] nums, int index, int sum, int S, Map<String, Integer> map){
String encodeString = index + "->" + sum;
if (map.containsKey(encodeString)){
return map.get(encodeString);
}
if (index == nums.length){
if (sum == S){
return 1;
}else {
return 0;
}
}
int curNum = nums[index];
int add = helper(nums, index + 1, sum - curNum, S, map);
int minus = helper(nums, index + 1, sum + curNum, S, map);
map.put(encodeString, add + minus);
return add + minus;
}
标签：记录 osi 递归 man help style public str for
原文地址：http://www.cnblogs.com/apanda009/p/7281704.html
